##### Class 12^{th} Chemistry Part I CBSE Solution

**Intext Questions Pg-98**- For the reaction R → P, the concentration of a reactant changes from 0.03M to 0.02M in 25…
- In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L-1 to 0.4 mol…

**Intext Questions Pg-103**- For a reaction, A + B → Product; the rate law is given by, r = k [A]1/2 [B]^2 . What is…
- The conversion of molecules X to Y follows second order kinetics. If concentration of X is…

**Intext Questions Pg-111**- A first order reaction has a rate constant 1.15 × 10-3 s-1. How long will 5 g of this…
- Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the…

**Intext Questions Pg-116**- What will be the effect of temperature on rate constant ?
- The rate of the chemical reaction doubles for an increase of 10K in absolute temperature…
- The activation energy for the reaction 2 HI(g) → H2 + I2 (g) is 209.5 kJ mol-1 at…

**Exercises**- From the rate expression for the following reactions, determine their order of reaction…
- For the reaction : 2A + B → A2B the Rate = k[A][B]^2 with k = 2.0 × 10-6 mol-2 L^2 sec-1.…
- The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of…
- The decomposition of dimethyl ether leads to formation of CH4, H2 and CO and the reaction…
- Mention the factors that affect the rate of chemical reaction?
- A reaction is second order with respect to a reactant. How is the rate of reaction…
- What is the effect of temperature on the rate constant of a reaction? How can this effect…
- In a pseudo first order hydrolysis of ester in water, the following results were obtained…
- A reaction is First Order in A and second order in B. (i) Write the differential rate…
- In a reaction between A and B, the initial rate of reaction(r0)was measured for different…
- The following results have been obtained during the kinetic studies of the reaction: 2A +…
- The reaction between A and B is first order with respect to A and zero order with respect…
- Calculate the half life of a first order reaction from their rate constants given below:…
- The half-life for radioactive decay of 14C is 5730 years. An archeological artifact…
- The experiemental data for decomposition of N2O5 [2N2O5→ 4NO2 + O2] in gas phase at 318 k…
- The rate constant for a first order reaction is 60 s-1. How much time will it take to…
- During nuclear explosion, one of the products is^90 Sr with half-life of 28.1 years. If…
- For a first order reaction, show that time required for 99% completion is twice the time…
- A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.…
- For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data…
- The following data were obtained during the first order thermal decomposition of SO2Cl2 at…
- The rate constant for the decomposition of N2O5 at various temperatures is given below:…
- The rate constant for the decomposition of hydrocarbons is 2.418 × 10-5s-1 at 546 K. If…
- Consider a certain reaction A → Products with k = 2.0 × 10-2s-1. Calculate the…
- Sucrose decomposes in acid solution into glucose and fructose according to the first order…
- The decomposition of hydrocarbon follows the equation k = (4.5 × 10^11 s-1) e-28000K/T.…
- The rate constant for the first order decomposition of H2O2 is given by the following…
- The decomposition of A into product has value of k as 4.5 × 10^3 s-1 at 10°C and energy of…
- The time required for 10% completion of a first order reaction at 298K is equal to that…
- The rate of a reaction quadruples when the temperature changes from 293 K to 313 K.…

**Intext Questions Pg-98**

- For the reaction R → P, the concentration of a reactant changes from 0.03M to 0.02M in 25…
- In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L-1 to 0.4 mol…

**Intext Questions Pg-103**

- For a reaction, A + B → Product; the rate law is given by, r = k [A]1/2 [B]^2 . What is…
- The conversion of molecules X to Y follows second order kinetics. If concentration of X is…

**Intext Questions Pg-111**

- A first order reaction has a rate constant 1.15 × 10-3 s-1. How long will 5 g of this…
- Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the…

**Intext Questions Pg-116**

- What will be the effect of temperature on rate constant ?
- The rate of the chemical reaction doubles for an increase of 10K in absolute temperature…
- The activation energy for the reaction 2 HI(g) → H2 + I2 (g) is 209.5 kJ mol-1 at…

**Exercises**

- From the rate expression for the following reactions, determine their order of reaction…
- For the reaction : 2A + B → A2B the Rate = k[A][B]^2 with k = 2.0 × 10-6 mol-2 L^2 sec-1.…
- The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of…
- The decomposition of dimethyl ether leads to formation of CH4, H2 and CO and the reaction…
- Mention the factors that affect the rate of chemical reaction?
- A reaction is second order with respect to a reactant. How is the rate of reaction…
- What is the effect of temperature on the rate constant of a reaction? How can this effect…
- In a pseudo first order hydrolysis of ester in water, the following results were obtained…
- A reaction is First Order in A and second order in B. (i) Write the differential rate…
- In a reaction between A and B, the initial rate of reaction(r0)was measured for different…
- The following results have been obtained during the kinetic studies of the reaction: 2A +…
- The reaction between A and B is first order with respect to A and zero order with respect…
- Calculate the half life of a first order reaction from their rate constants given below:…
- The half-life for radioactive decay of 14C is 5730 years. An archeological artifact…
- The experiemental data for decomposition of N2O5 [2N2O5→ 4NO2 + O2] in gas phase at 318 k…
- The rate constant for a first order reaction is 60 s-1. How much time will it take to…
- During nuclear explosion, one of the products is^90 Sr with half-life of 28.1 years. If…
- For a first order reaction, show that time required for 99% completion is twice the time…
- A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.…
- For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data…
- The following data were obtained during the first order thermal decomposition of SO2Cl2 at…
- The rate constant for the decomposition of N2O5 at various temperatures is given below:…
- The rate constant for the decomposition of hydrocarbons is 2.418 × 10-5s-1 at 546 K. If…
- Consider a certain reaction A → Products with k = 2.0 × 10-2s-1. Calculate the…
- Sucrose decomposes in acid solution into glucose and fructose according to the first order…
- The decomposition of hydrocarbon follows the equation k = (4.5 × 10^11 s-1) e-28000K/T.…
- The rate constant for the first order decomposition of H2O2 is given by the following…
- The decomposition of A into product has value of k as 4.5 × 10^3 s-1 at 10°C and energy of…
- The time required for 10% completion of a first order reaction at 298K is equal to that…
- The rate of a reaction quadruples when the temperature changes from 293 K to 313 K.…

###### Intext Questions Pg-98

**Question 1.**For the reaction R → P, the concentration of a reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

**Answer:**Given-

Initial concentration(R_{1}) = 0.03M

Final concentration(R_{2}) = 0.02M

Time = 25 mins.

The formula for average rate of the reaction is,

→ Equation 1

∵ {R} = (R_{2})-( R_{1}), the equation 1 is written as,

= 4 × 10^{-4} mol L^{-1} min^{-1}

The average rate of reaction in seconds is given by,

=

(dividing by 60 to convert minutes to seconds)

= 6.6 × 10^{-6} mol L^{-1} s^{-1}

__The average rate of the reaction in minutes is____4 × 10__^{-4} mol L^{-1} min^{-1} and in seconds is 6.6 × 10^{-6} mol L^{-1} s^{-1}

**Question 2.**In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L^{–}^{1} to 0.4 mol L^{–}^{1} in 10 minutes. Calculate the rate during this interval?

**Answer:**Given-

Initial concentration(A_{1}) = 0.5M

Final concentration(A_{2}) = 0.4M

Time = 10 mins.

The formula for average rate of the reaction is,

→ Equation 1

∵ {A} = (A_{2})-( A_{1}), the equation 1 is written as,

=

= 0.005 mol L^{-1} min^{-1}

= 5 × 10^{-3} mol L^{-1} min^{-1}

__The average rate of the reaction is____5 × 10__^{-3} mol L^{-1} min^{-1}

**Question 1.**

For the reaction R → P, the concentration of a reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

**Answer:**

Given-

Initial concentration(R_{1}) = 0.03M

Final concentration(R_{2}) = 0.02M

Time = 25 mins.

The formula for average rate of the reaction is,

→ Equation 1

∵ {R} = (R_{2})-( R_{1}), the equation 1 is written as,

= 4 × 10^{-4} mol L^{-1} min^{-1}

The average rate of reaction in seconds is given by,

=

(dividing by 60 to convert minutes to seconds)

= 6.6 × 10^{-6} mol L^{-1} s^{-1}

__The average rate of the reaction in minutes is____4 × 10 ^{-4} mol L^{-1} min^{-1} and in seconds is 6.6 × 10^{-6} mol L^{-1} s^{-1}__

**Question 2.**

In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L^{–}^{1} to 0.4 mol L^{–}^{1} in 10 minutes. Calculate the rate during this interval?

**Answer:**

Given-

Initial concentration(A_{1}) = 0.5M

Final concentration(A_{2}) = 0.4M

Time = 10 mins.

The formula for average rate of the reaction is,

→ Equation 1

∵ {A} = (A_{2})-( A_{1}), the equation 1 is written as,

=

= 0.005 mol L^{-1} min^{-1}

= 5 × 10^{-3} mol L^{-1} min^{-1}

__The average rate of the reaction is____5 × 10 ^{-3} mol L^{-1} min^{-1}__

###### Intext Questions Pg-103

**Question 1.**For a reaction, A + B → Product; the rate law is given by, r = k [ A]^{1/2} [B]^{2}. What is the order of the reaction?

**Answer:**The order of the reaction is sum of the powers of concentration of reactants in the rate law. According to this,

The order of the reaction = 1/2 + 2

=

__The order of the reaction is 2.5__

**Question 2.**The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y ?

**Answer:**Let the reaction be X→Y

As, this reaction follows second order kinetics, the rate of the reaction will be,

Rate = k(X)^{2} → Equation 1

Since the concentration of X is increased three times, Equation 1

will become,

Rate = k(3X)^{2}

= k × 9(X)^{2}

∴ The rate of formation of Y will become 9 times.

__Thus, the rate of formation of Y will become 9 times__

**Question 1.**

For a reaction, A + B → Product; the rate law is given by, r = k [ A]^{1/2} [B]^{2}. What is the order of the reaction?

**Answer:**

The order of the reaction is sum of the powers of concentration of reactants in the rate law. According to this,

The order of the reaction = 1/2 + 2

=

__The order of the reaction is 2.5__

**Question 2.**

The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y ?

**Answer:**

Let the reaction be X→Y

As, this reaction follows second order kinetics, the rate of the reaction will be,

Rate = k(X)^{2} → Equation 1

Since the concentration of X is increased three times, Equation 1

will become,

Rate = k(3X)^{2}

= k × 9(X)^{2}

∴ The rate of formation of Y will become 9 times.

__Thus, the rate of formation of Y will become 9 times__

###### Intext Questions Pg-111

**Question 1.**A first order reaction has a rate constant 1.15 × 10^{-3} s^{-1}. How long will 5 g of this reactant take to reduce to 3 g?

**Answer:**Given-

Rate constant, k = 1.15 × 10^{-3} s^{-1}

Initial quantity, R_{0} = 5 g

Final quantity, R = 3 g

According to the formula of first order reaction,

t =

t = 443.8 s

t = 4.438 × 10^{2} s

__The time taken for 5g of the reactant to reduce to 3 g is 4.438 × 10__^{2} s

**Question 2.**Time required to decompose SO_{2}Cl_{2} to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

**Answer:**Given-

60 mins

Using the formula for half life, , we get,

∴ k = 1.155 × 10^{-2} min^{-1}

__The rate constant of the reaction, k is____1.155 × 10__^{-2} min^{-1}

**Question 1.**

A first order reaction has a rate constant 1.15 × 10^{-3} s^{-1}. How long will 5 g of this reactant take to reduce to 3 g?

**Answer:**

Given-

Rate constant, k = 1.15 × 10^{-3} s^{-1}

Initial quantity, R_{0} = 5 g

Final quantity, R = 3 g

According to the formula of first order reaction,

t =

t = 443.8 s

t = 4.438 × 10^{2} s

__The time taken for 5g of the reactant to reduce to 3 g is 4.438 × 10 ^{2} s__

**Question 2.**

Time required to decompose SO_{2}Cl_{2} to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

**Answer:**

Given-

60 mins

Using the formula for half life, , we get,

∴ k = 1.155 × 10^{-2} min^{-1}

__The rate constant of the reaction, k is____1.155 × 10 ^{-2} min^{-1}__

###### Intext Questions Pg-116

**Question 1.**What will be the effect of temperature on rate constant ?

**Answer:**The rate constant of a chemical reaction normally increases with increase in temperature. It is observed that for a chemical reaction with rise in temperature by 10°C, the rate constant is nearly doubled and this temperature dependence of the rate of a chemical reaction can

be accurately explained by Arrhenius equation,

k = A

__Thus, the rate constant of a chemical reaction normally increases with increase in temperature.__

**Question 2.**The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate E_{a}.

**Answer:**Given-

Initial temperature, T_{1} = 298 K

Final temperature, T_{2} = 298 K + 10 K = 308 K

Knowing that the rate constant of a chemical reaction normally increases with increase in temperature, we assume that,

Initial value of rate constant, k_{1} = k

Final value of rate constant, k_{2} = 2k

Using Arrhenius equation,

→ Equation 1

where, R = 8.314 J K^{-1} mol^{-1} (gas constant).

Substituting all the values in equation 1, we get,

log 2 =

E_{a} =

E_{a} =

E_{a} = 52897 J mol^{-1}

E_{a} = 52.897 kJ mol^{-1}

__The energy of activation, E___{a} is 52.897 kJ mol^{-1}

**Question 3.**The activation energy for the reaction 2 HI(g) → H_{2} + I_{2} (g) is 209.5 kJ mol^{–}^{1} at 581K.Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

**Answer:**Given-

Energy of activation, E_{a}= 209.5 kJ mol^{–}^{1}

E_{a} in joules = 209.5 × 1000 = 209500 J mol^{–}^{1}

Temperature, T = 581K

Gas constant, R = 8.314 J K^{-1} mol^{-1}

Using Arrhenius equation,

k = A

In this equation, the term represents the number of molecules which have kinetic energy greater than the activation energy, E_{a}.

∴ The number of molecules = → Equation 1

Substituting all the known values in equation 1, we get,

⇒

= e^{-43.3708}

Finding the value of anti ln(43.3708), we get, 1.47 × 10^{-19}

__The fraction of molecules of reactants having energy equal to or greater than activation energy is____1.47 × 10__^{-19}

**Question 1.**

What will be the effect of temperature on rate constant ?

**Answer:**

The rate constant of a chemical reaction normally increases with increase in temperature. It is observed that for a chemical reaction with rise in temperature by 10°C, the rate constant is nearly doubled and this temperature dependence of the rate of a chemical reaction can

be accurately explained by Arrhenius equation,

k = A

__Thus, the rate constant of a chemical reaction normally increases with increase in temperature.__

**Question 2.**

The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate E_{a}.

**Answer:**

Given-

Initial temperature, T_{1} = 298 K

Final temperature, T_{2} = 298 K + 10 K = 308 K

Knowing that the rate constant of a chemical reaction normally increases with increase in temperature, we assume that,

Initial value of rate constant, k_{1} = k

Final value of rate constant, k_{2} = 2k

Using Arrhenius equation,

→ Equation 1

where, R = 8.314 J K^{-1} mol^{-1} (gas constant).

Substituting all the values in equation 1, we get,

log 2 =

E_{a} =

E_{a} =

E_{a} = 52897 J mol^{-1}

E_{a} = 52.897 kJ mol^{-1}

__The energy of activation, E _{a} is 52.897 kJ mol^{-1}__

**Question 3.**

The activation energy for the reaction 2 HI(g) → H_{2} + I_{2} (g) is 209.5 kJ mol^{–}^{1} at 581K.Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

**Answer:**

Given-

Energy of activation, E_{a}= 209.5 kJ mol^{–}^{1}

E_{a} in joules = 209.5 × 1000 = 209500 J mol^{–}^{1}

Temperature, T = 581K

Gas constant, R = 8.314 J K^{-1} mol^{-1}

Using Arrhenius equation,

k = A

In this equation, the term represents the number of molecules which have kinetic energy greater than the activation energy, E_{a}.

∴ The number of molecules = → Equation 1

Substituting all the known values in equation 1, we get,

⇒

= e^{-43.3708}

Finding the value of anti ln(43.3708), we get, 1.47 × 10^{-19}

__The fraction of molecules of reactants having energy equal to or greater than activation energy is____1.47 × 10 ^{-19}__

###### Exercises

**Question 1.**From the rate expression for the following reactions, determine their order of reaction and dimensions of the rate constants.

(i) 3NO(g) → N_{2}O (g) Rate = k[NO]^{2}

(ii) H_{2}O_{2} (aq) + 3I^{–}(aq) + 2H^{+}→ 2H_{2}O(l) + I^{–}_{3} Rate = k[H_{2}O_{2}][I^{–}]

(iii) CH_{3}CHO(g) → CH_{4}(g) + CO(g) Rate = k[CH_{3}CHO]^{3/2}

(iv) C_{2}H_{5}Cl (g) → C_{2}H_{4}(g) + HCl (g) Rate = k[C_{2}H_{5}Cl]

Order is sum of all powers of the concentrations of the reactant in “rate law expression”

**Answer:**(i) 3NO(g) → N_{2}O(g)

Rate = k[NO]^{2}

Here rate law is dependent on concentration of NO only.And power raised to concentration of NO is 2. Hence __order of reaction is 2.__

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in . So, unit of K will be

i.e. [concentration]^{-1}[ time]^{-1} . If concentration is in mol L^{-1} and time is in sec. then __dimensions of k will be__ : __mol__^{-1} L sec^{-1}

(ii) H_{2}O_{2}(aq) + 3I^{-}(aq) + 2H^{+} → 2 H_{2}O(l) + I_{3}^{-}

Rate = k[H_{2}O_{2}][I^{-}]

Here rate law is dependent on concentration of H_{2}O_{2} and I^{-}. And power raised to concentration of H_{2}O_{2} is 1, and power raised to concentration of I^{-} is 1. order is sum of powers hence, __order of given reaction is 2.__

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in [concentration/time]. So, unit of K will be [Concentration]/[time][concentration]^{2} i.e. [concentration]^{-1}[ time]^{-1}. If concentration is in mol L^{-1} and time is in sec then __dimensions of k will be____mol__^{-1} L sec^{-1}

(iii) CH_{3}CHO(g) → CH_{4}(g) + CO(g)

Rate = k[CH_{3}CHO]^{3/2}

Here rate law is dependent on concentration of CH_{3}CHO.And power raised to CH_{3}CHO is 3/2. Hence order of reaction is 3/2.

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in [concentration/time]. So, unit of K will be [Concentration]/[time][concentration]^{3/2} i.e. [concentration]^{-1/2}[time]^{-1}.

If concentration is in mol L^{-1} and time is in sec. Then __dimensions of K will be mol__^{-1/2}L^{1/2} sec^{-1}

(iv) C_{2}H_{5}Cl (g) → C_{2}H_{4}(g) + HCl(g)

Rate = k[C_{2}H_{5}Cl]

Here rate law is dependent on concentration of C_{2}H_{5}Cl. And power raised to C_{2}H_{5}Cl is 1. Hence order of reaction is 1.

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in [concentration/time]. So, unit of K will be [Concentration]/[time][concentration]^{1}. i.e. [time]^{-1}

If concentration is in mol L^{-1} and time is in sec. Then __dimensions of K will be sec__^{-1}.

**Question 2.**For the reaction :

2A + B → A_{2}B

the Rate = k[A][B]^{2} with k = 2.0 × 10^{-6} mol^{-2} L^{2} sec^{-1}. Calculate the initial rate of the reaction when [A] = 0.1 mol L^{-1} and [B] = 0.2 mol L^{-1}. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^{-1}.

**Answer:**a) Rate = k[A][B]^{2}, Rate = 2.0 × 10^{-6}[0.1][0.2]^{2}

Rate = 8 × 10^{-9} mol L^{-1}sec^{-1}

b)

Situation when A is remained 0.06 mol L^{-1}

Now, According to rate law, Rate = k[A][B]^{2}

Rate = 2 × 10^{-6}[0.06][0.18]^{2}

i.e. Rate = 3.888 × 10^{-9} mol L^{-1}sec^{-1}

__Initial rate of reaction is 8 × 10__^{-9} mol L^{-1}sec^{-1}. and rate when concentration of A is 0.06 mol L^{-1} is 3.888 × 10^{-9} mol L^{-1}sec^{-1}.

**Question 3.**The decomposition of NH_{3} on platinum surface is zero order reaction.

What are the rates of production of N_{2} and H_{2}. if k = 2.5 × 10^{-4} mol L^{-1} sec^{-1}?

**Answer:**2NH_{3}(g) → N_{2}(g) + 3H_{2}(g)

Rate of zero order reaction is equal to rate constant. i.e. Rate = 2.5 × 10^{-4}mol L^{-1}sec^{-1}.

According to rate law,

2.5 × 10^{-4}mol L^{-1}sec^{-1} =

i.e. the rate of production of N_{2} is 2.5 × 10^{-4}mol L^{-1} sec^{-1}.

According to rate law,

i.e. rate of formation of H_{2} is 3 times rate of reaction = 3 × 2.5 × 10^{-4}mol L^{-1}sec^{-1}

= 7.5 × 10^{-4}mol L^{-1}sec^{-1}

__Rate of formation of N___{2} and H_{2} is 2.5 × 10^{-4} mol L^{-1}sec^{-1} and 7.5 × 10^{-4} mol L^{-1}sec^{-1} respectively.

**Question 4.**The decomposition of dimethyl ether leads to formation of CH_{4}, H_{2} and CO and the reaction rate is given by,

Rate = k[CH_{3}OCH_{3}]^{3/2}. The rate of reaction is followed by an increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether i.e. Rate = k[pCH_{3}OCH_{3}]^{3/2}. If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

**Answer:**Rate of given chemical reaction will be represented as

.

Hence units of rate is bar min^{-1}

To find units of K, K = rate/[pCH_{3}OCH_{3}]^{3/2}

The unit of k = bar ^{-1/2}min^{-1}.

__Units of Rate- bar min__^{-1}. and Units of Rate constant K : bar ^{-1/2}min ^{-1}

**Question 5.**Mention the factors that affect the rate of chemical reaction?

**Answer:**Factors affecting rate of reaction-

1) Temperature (increasing temperature increases rate of reaction)

2) Concentration or pressure of reactants. (Increasing concentration or pressure increases rate of reaction)

3) Presence or absence of a catalyst. (Adding catalyst mostly increases the rate of reaction)

4) The surface area of solid reactant. (If reaction is processing over solid reactant then increasing its surface area increases rate of reaction)

5) Nature of reactants

**Question 6.**A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration is

(i) doubled (ii) reduced to half

**Answer:**Let suppose reaction

A ⇒ B;

having rate law, Rate = K [A]^{2}

(i) If the concentration of A is doubled then the rate will affect by the square of concentration i.e. rate will become 2^{2} = 4 times.

(ii) If the concentration of A is halved then the rate will affect the square of concentration i.e (1/2)^{2} = 1/4 times.

__(i) Rate becomes 4 times of initial Rate (ii) Rate becomes 1/4 times of initial Rate__

**Question 7.**What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?

**Answer:**__Increase in temperature increases the rate constant of a reaction__. as we know increase in temperature increases the rate of reaction to satisfy the equation Rate = k [concentration]^{n} where n can be any real number. k have to increase as concentration is almost not changing over small temperature change. Increasing temperature by 10^{0}C almost double the rate constant.

This can be represented quantitively by the help of arrehinus equation-

__K = Ae__^{-Ea/RT}, where k is rate constant, E_{a} is activation energy, R is universal gas constant, T is absolute temperature.

**Question 8.**In a pseudo first order hydrolysis of ester in water, the following results were obtained

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds

(ii) Calculate the pseudo first-order rate constant for the hydrolysis of the ester.

**Answer:**(i) Average rate of reaction over interval is [change in concentration]/[time taken] i.e.

(ii) the pseudo first-order rate constant can be calculated by

K = (2.303/t) log(C_{i}/C_{t}) where K is Rate constant,

t is time taken,

C_{i} is initial concentration

C_{t} is Concentration at time t.

K = (2.303/30) log (0.55/0.31)

⇒ K = 1.9 × 10^{-2} sec^{-1}

__(i) Average rate between 30 to 60 sec is 0.00467 mol L__^{-1}sec^{-1}

__(ii) Pseudo first order rate constant is 1. × 10__^{-2}sec^{-1}

**Question 9.**A reaction is First Order in A and second order in B.

(i) Write the differential rate equation.

(ii) How is the rate affected on increasing the concentration of B three times?

(iii) How is the rate affected when the concentrations of both A and B are doubled?

**Answer:**(i) Order is power raised to reactant in rate law, hence,

Rate = k[A][B]^{2}

(ii) When the concentration of B is increased three times then the rate is affected by the square of reactant. The rate is increased by 9 Times.

(iii) When concentration of reactant both A and B is doubled then the rate will have affected as square of reactant B and Two times of Reactant A. Overall increase in rate is 8 times

__(i) When the concentration of B is increased by three times, the rate is increased by nine times.__

__(ii) When conc. of both reactants doubled then Rate increased 8 times.__

**Question 10.**In a reaction between A and B, the initial rate of reaction(r_{0})was measured for different initial concentrations of A and B are given below:

What is the order of reaction with respect to A and B?

**Answer:**When concentration of B is changed then rate of reaction doesn't change that means order with respect to B is 0. But when the concentration of A is doubled rate increased by 2.82 times i.e.2^{1.5} = 2.82.Hence order with respect to A is 1.5.

Order with respect to A and B is 1.5 and 0 respectively.

**Question 11.**The following results have been obtained during the kinetic studies of the reaction:

2A + B → C + D

Determine the Rate law and the rate constant for the reaction.

**Answer:**By comparing Experiment I and IV if we increase the concentration of A by 4 times then Rate also increased by 4 times. That means order with respect to A is 1.

By comparing Experiment II and III if we double the concentration of B Rate increases by 4 times that means order with respect to B is 2.

Rate law of reaction will be, Rate = k [A][B]^{2}

To find K, K = rate/[A][B]^{2} i.e. K = 6.0 × 10^{-3}/[0.1][0.1]^{2}

K = 6 mol^{-2}L^{2}sec^{-1}

__Order with respect to A and B is 1 and 2 Respectively. And value of K(rate constant) is = 6 mol__^{-2}L^{2}sec^{-1}

**Question 12.**The reaction between A and B is first order with respect to A and zero order with respect to B.Phil. in the blanks in the following table:

**Answer:**As reaction is first order with respect to A and zero Order with respect to B. Then changing the concentration of B won’t affect the rate of reaction and increasing concentration of A ‘n’ times will increase the rate by ‘n’ times. By this logic lets fill the table- In first blank space concentration of A will be 0.2 mol L^{-1} because the rate is doubled. In second blank space, Rate will be 8 × 10^{-2}mol L^{-1}min^{-1} because the concentration of A is increased 4 Times. In third blank space concentration of A will be 0.1 mol L^{-1} because the rate is same as in experiment I.

**Question 13.**Calculate the half life of a first order reaction from their rate constants given below:

(i) 200 s^{-1} (ii) 2 min^{-1} (iii) 4 years^{-1}

**Answer:**Half life of first order reaction is, t_{1/2} = ln2/K where t_{1/2} is half life of first order reaction, K is rate constant of First order reaction.

(i) t_{1/2} = ln2/200 s^{-1}

⇒ t_{1/2} = 0.693/200 s^{-1} (∵ ln2 = 0.693)

⇒ t_{1/2} = 0.003465 sec.

(ii) t_{1/2} = ln2/2 min^{-1}

⇒ t_{1/2} = 0.693/2 min^{-1} (∵ ln2 = 0.693)

⇒ t_{1/2} = 0.3465 min

(iii) t_{1/2} = ln2/4 year^{-1}

⇒ t_{1/2} = 0.693/4 years ^{-1} (∵ ln2 = 0.693)

⇒ t_{1/2} = 0.17325 year.

__Half life of 3 reactions are 0.003465 sec, 0.3465 min, 0.17325 year respectively.__

**Question 14.**The half-life for radioactive decay of 14C is 5730 years. An archeological artifact containing wood had only 80%of the 14C found in a living tree. Estimate the age of the sample.

**Answer:**Radio active decay occurs via first order rate law,

t_{1/2} = 5730 years. rate constant(k) of given decay is 0.693/t_{1/2}

⇒ 0.693/5730 = 1.2 × 10^{-4} year^{-1}

By first order integrated rate law age of sample will be,

where T is the age of sample A_{0} is the initial activity of the sample. and A_{t} is the activity of sample at any time t.

T = 0.18 × 10^{4} years.

__Age of given sample is 0.18 × 10__^{4} years.

**Question 15.**The experiemental data for decomposition of N_{2}O_{5}

[2N_{2}O_{5}→ 4NO_{2} + O_{2}]

in gas phase at 318 k are given below:

(i) plot N_{2}O_{5} against t

(ii) Find the half-life period for the Reaction.

(iii) Draw a graph between log[N_{2}O_{5}]and t.

(iv) what is the rate law?

(v) Calculate the rate constant.

(vi) Calculate the half-life period from k and compare it with (ii)

**Answer:**

(i)

(ii) initial conc.is 1.63 × 10^{-2} so half conc. is 0.815 × 10^{-2}M From given information and graph t_{1/2} is 1440 sec.

(iii)

(iv) As log [N_{2}O_{5}] vs time is straight line given reaction is first order. Hence its rate law will be, Rate = k[N_{2}O_{5}]

(v) The slope of above graph is slope = 0.000209

K = 2.303 × slope

⇒ 4.82 × 10^{-4}sec^{-1}

Now, t_{1/2} = 0.693/K.

⇒ 0.693/4.82 × 10^{-4}

⇒ t_{1/2} = 1438 sec. which is almost equal to (ii)

**Question 16.**The rate constant for a first order reaction is 60 s^{–1}. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

**Answer:**Given:

Order of the reaction = 1

Let, Initial concentration [R]^{°}= x

Final concentration [R] = x/16

Rate constant k = 60 s^{-1}

We know, time

⇒

⇒

Solving, we get __t = 4.6 × 10__^{-2}s

**Question 17.**During nuclear explosion, one of the products is ^{90}Sr with half-life of 28.1 years. If 1Î¼g of ^{90}Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

**Answer:**Initial concentration, [R]_{°} = 1Î¼g

Final concentration, [R]

Half-life t_{1/2} = 28.1 yrs

Solution:

We know, t_{1/2} = 0.693/k

Where, k – rate constant

⇒ k = 0.693/ t_{1/2}

⇒ k = 0.693/(28.1 yrs)

⇒ k = 0.0246 yrs^{-1}

Also,

If t = 10yrs, then, using the formula, we get,

⇒

⇒

⇒

⇒

⇒

⇒

⇒ = 0.7824Î¼g

If t = 60yrs, then again, we get,

⇒

⇒

⇒

⇒

⇒ [R]_{60} = 0.228 Î¼g

∴ 0.2278Î¼g of ^{90}Sr will be left after 60 years.

**Question 18.**For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

**Answer:**Let, initial concentration be [R]_{°}

Concentration at 90% completion be ((100-90)/100)×[R]_{°}

∴ Concentration at 90% be 0.1[R]_{°}

Concentration at 99% completion be ((100-99)/100)× [R]_{°}∴ Concentration at 99% be 0.01[R]_{°}

We know, time

Time taken for 90% completion is

⇒

⇒

Time taken for 99% completion is

⇒

⇒

⇒ t_{99} = 2t_{90}

Hence, the time taken to complete 9% of the first order reaction is twice the time required for the completion of 90% of the reaction.

**Question 19.**A first order reaction takes 40 min for 30% decomposition. Calculate t_{1/2}.

**Answer:**Given:

Time t = 40 min

When 30% decomposition is undergone, 70% is the concentration.

We know, time taken

Where, k- rate constant

[R]_{°} -Initial concentration

[R]-Concentration at time ‘t’

⇒

⇒

⇒

⇒ 258.23 = (2.303/k)

We know, Half-life t_{1/2} = 0.693/k

Which can be written as, t_{1/2} = 0.3010 × (2.303/k)

⇒ t_{1/2} = 0.3010 × 258.23

⇒ t_{1/2} = 77.72 min

**Question 20.**For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

Calculate the rate constant.

**Answer:**When t = 0, the total partial pressure is P_{0} = 35.0 mm of Hg

When time t = t, the total partial pressure is P_{t} = P_{0} + p

P_{0}-p = P_{t}-2p, but by the above equation, we know p = P_{t}-P_{0}

Hence, P_{0}-p = P_{t}-2( P_{t}-P_{0})

Thus, P_{0}-p = 2P_{0} – P_{t}

We know that time

Where, k- rate constant

[R]_{°} -Initial concentration of reactant

[R]-Concentration of reactant at time ‘t’

Here concentration can be replaced by the corresponding partial pressures.

Hence, the equation becomes,

→ equation 1

At time t = 360 s, P_{t} = 54 mm of Hg and P_{0} = 30 mm of Hg,

Substituting in equation 1,

⇒

⇒ k = 2.175 × 10^{-3} s^{-1}

At time t = 720 s, P_{t} = 63 mm of Hg and P_{0} = 30 mm of Hg,

Substituting in equation 1,

⇒

Thus, k = 2.235 × 10^{-3} s^{–1}

Taking average, k = (2.235 × 10^{-3} s^{–1} + 2.175 × 10^{-3} s^{–1})/2

∴ k = 2.21 × 10^{-3} s^{–1}.

**Question 21.**The following data were obtained during the first order thermal decomposition of SO_{2}Cl_{2} at a constant volume.

SO_{2}Cl_{2}_{(g)}→ SO_{2 (g)} + Cl_{2 (g)}

Calculate the rate of the reaction when total pressure is 0.65 atm

**Answer:**When t = 0, the total partial pressure is P_{0} = 0.5 atm

When time t = t, the total partial pressure is P_{t} = P_{0} + p

P_{0}-p = P_{t}-2p, but by the above equation, we know p = P_{t}-P_{0}

Hence, P_{0}-p = P_{t}-2(P_{t}-P_{0})

Thus, P_{0}-p = 2P_{0} – P_{t}

We know that time

Where, k- rate constant

[R]_{°} -Initial concentration of reactant

[R]-Concentration of reactant at time ‘t’

Here concentration can be replaced by the corresponding partial pressures.

Hence, the equation becomes,

⇒ → equation 1

At time t = 100 s, P_{t} = 0.6 atm and P_{0} = 0.5 atm,

Substituting in equation 1,

⇒

Thus, k = 2.231 × 10^{-3} s^{-1}

The rate of reaction R = k × P_{S02Cl2}

When total pressure P_{t} = 0.65 atm and P_{0} = 0.5 atm, then

P_{S02Cl2} = 2P_{0}-P_{t}

Thus, substituting the values, P_{S02Cl2} = 2(0.5)-0.6 = 0.35 atm

R = k × P_{S02Cl2} = 2.231 × 10^{-3} s^{–1} × 0.35

Rate of the reaction R = 7.8 × 10^{-4}atm s^{–1}

**Question 22.**The rate constant for the decomposition of N_{2}O_{5} at various temperatures is given below:

Draw a graph between ln k and 1/T and calculate the values of A and Ea. Predict the rate constant at 30° and 50°C.

**Answer:**To convert the temperature in °C to °K we add 273 K.

The graph is given as:

The Arrhenius equation is given by k = Ae^{-Ea/RT}

Where, k- Rate constant

A- Constant

E_{a}-Activation Energy

R- Gas constant

T-Temperature

Taking natural log on both sides,

ln k = ln A-(E_{a}/RT)

→ equation 1

By plotting a graph, ln K Vs 1/T, we get y-intercept as ln A and Slope is –E_{a}/R.

Slope = (y_{2}-y_{1})/(x_{2}-x_{1})

By substituting the values, slope = -12.301

⇒ –E_{a}/R = -12.301

But, R = 8.314 JK^{-1}mol^{-1}

⇒ E_{a} = 8.314 JK^{-1}mol^{-1} × 12.301 K

⇒ E_{a} = 102.27 kJ mol^{-1}

Substituting the values in equation 1 for data at T = 273K

⇒

⇒

(∵ At T = 273K, ln k = -7.147)

On solving, we get ln A = 37.911

∴ A = 2.91×10^{6}

When T = 30^{0}C, hence T = 30 + 273 = 303K

⇒

⇒

⇒ ln k = -2.8

⇒ k = 6.08x10^{-2}s^{-1}.

When T = 50^{0}C, hence T = 50 + 273 = 323K

Substituting in equation 1,

A = 2.91 × 10^{6}, E_{a} = 102.27 kJ mol^{-1}

Ln K = ln(2.91 × 10^{6})- 102.27/(8.314 × 323)

Thus, ln k = -0.5 and k = 0.607s^{-1}.

**Question 23.**The rate constant for the decomposition of hydrocarbons is 2.418 × 10^{–5}s^{–1} at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

**Answer:**Given,

k = 2.418 × 10^{-5} s^{-1}

T = 546 K

E_{a} = 179.9 kJ mol^{-1} = 179.9 × 10^{3}J mol^{-1}

The Arrhenius equation is given by k = Ae^{-Ea/RT}

Taking natural log on both sides,

Ln k = ln A-(E_{a}/RT)

Substituting the values,

ln(2.418 × 10^{-5} ) = ln A-179.9/(8.314 × 546)

ln A = 12.5917

A = 3.9 × 10^{12} s^{-1}(approximately)

**Question 24.**Consider a certain reaction A → Products with k = 2.0 × 10^{–}^{2}s^{-}^{1}. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L^{–1}.

**Answer:**Given,

k = 2.0 × 10^{–}^{2}s^{-}^{1}

time t = 100s

Concentration [A_{0}] = 1.0 mol L^{-1}

We know,

On substituting the values,

Log(1/[A]) = 2.303/2

Log[A] = -2.303/2

[A] = 0.135 mol L^{–1}

**Question 25.**Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_{1/2} = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

**Answer:**t_{1/2} = 3.00 hours

We know, t_{1/2} = 0.693/k

∴ k = 0.693/3

k = 0.231hrs^{-1}

We know, time

Where, k- rate constant

[R]_{°} -Initial concentration

[R]-Concentration at time ‘t’

Thus, substituting the values,

log([R]_{0}/[R]) = 0.8

log([R]/[R]_{0}) = -0.8

[R]/[R]_{0} = 0.158

Hence, 0.158 fraction of sucrose remains.

**Question 26.**The decomposition of hydrocarbon follows the equation

k = (4.5 × 10^{11}s^{–1}) e^{-28000K}/T. Calculate E_{a}.

**Answer:**The given equation is

k = (4.5 x 10^{11} s^{-1}) e^{-28000 K/T} (i)

Comparing, Arrhenius equation

k = Ae ^{-E}_{a}^{/RT} (ii)

We get, E_{a} / RT = 28000K / T

⇒ E_{a} = R x 28000K

= 8.314 J K^{-1}mol^{-1} × 28000 K

= 232792 J mol^{–1} or 232.792 kJ mol^{–1}

**Question 27.**The rate constant for the first order decomposition of H_{2}O_{2} is given by the following equation: log k = 14.34 – 1.25 × 10^{4}K/T

Calculate E_{a} for this reaction and at what temperature will its half-period be 256 minutes?

**Answer:**We know, The Arrhenius equation is given by k = Ae^{-Ea/RT}

Taking natural log on both sides,

Ln k = ln A-(E_{a}/RT)

Thus, log k = log A -(E_{a}/2.303RT) → eqn 1

The given equation is log k = 14.34 – 1.25 × 10^{4}K/T → eqn 2

Comparing 2 equations,

E_{a}/2.303R = 1.25 × 10^{4}K

E_{a} = 1.25 × 10^{4}K × 2.303 × 8.314

E_{a} = 239339.3 J mol^{-1} (approximately)

E_{a} = 239.34 kJ mol^{-1}

Also, when t_{1/2} = 256 minutes,

k = 0.693 / t_{1/2}

= 0.693 / 256

= 2.707 × 10^{-3} min^{-1}

k = 4.51 × 10^{-5}s^{–1}

Substitute k = 4.51 × 10^{-5}s^{–1} in eqn 2,

log 4.51 × 10^{-5} s^{–1} = 14.34 – 1.25 × 10^{4}K/T

log(0.654-5) = 14.34– 1.25 × 10^{4}K/T

T = 1.25 × 10^{4}/[ 14.34- log(0.654-5)]

T = 668.9K or T = 669 K

**Question 28.**The decomposition of A into product has value of k as 4.5 × 10^{3} s^{–1} at 10°C and energy of activation 60 kJ mol^{–1}. At what temperature would k be 1.5 × 10^{4}s^{–1}?

**Answer:**From Arrhenius equation, we obtain

Also, k_{1} = 4.5 × 10^{3} s ^{- 1}

T_{1} = 273 + 10 = 283 K

k_{2} = 1.5 × 10^{4} s ^{- 1}

E_{a} = 60 kJ mol ^{- 1} = 6.0 × 10^{4} J mol ^{- 1}

Then,

→ 0.5229 = 3133.627 × (T_{2}-283)/(283 × T_{2})

→ 0.0472T_{2} = T_{2}-283

T_{2} = 297K or T_{2} = 24^{0} C

**Question 29.**The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 10^{10}s^{–1}. Calculate k at 318K and E_{a}.

**Answer:**We know, time t = (2.303/k) × log([R]_{0}/[R])

Where, k- rate constant

[R]_{°} -Initial concentration

[R]-Concentration at time ‘t’

At 298K, If 10% is completed, then 90% is remaining.

t = (2.303/k) × log ([R]_{0}/0.9[R]_{0})

t = (2.303/k) × log (1/0.9)

t = 0.1054 / k

At temperature 308K, 25% is completed, 75% is remaining

t’ = (2.303/k’) × log ([R]_{0}/0.75[R]_{0})

t’ = (2.303/k’) × log (1/0.75)

t’ = 2.2877 / k'

But, t = t’

0.1054 / k = 2.2877 / k'

k' / k = 2.7296

From Arrhenius equation, we obtain

log k_{2}/k_{1} = (E_{a} / 2.303 R) × (T_{2} - T_{1}) / T_{1}T_{2}

Substituting the values,

⇒

E_{a} = 76640.09 J mol^{-1} or 76.64 kJ mol^{-1}

We know, log k = log A –E_{a}/RT

Log k = log(4 × 10^{10})-(76.64kJ mol^{-1}/(8.314 × 318))

Log k = -1.986

∴ k = 1.034 x 10^{-2} s ^{-1}

**Question 30.**The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

**Answer:**Given, k_{2} = 4k_{1}, T_{1} = 293K and T_{2} = 313K

We know, From Arrhenius equation, we obtain

⇒

⇒

On solving we get,

E_{a} = 58263.33 J mol^{-1} or 58.26 kJ mol^{-1}

**Question 1.**

From the rate expression for the following reactions, determine their order of reaction and dimensions of the rate constants.

(i) 3NO(g) → N_{2}O (g) Rate = k[NO]^{2}

(ii) H_{2}O_{2} (aq) + 3I^{–}(aq) + 2H^{+}→ 2H_{2}O(l) + I^{–}_{3} Rate = k[H_{2}O_{2}][I^{–}]

(iii) CH_{3}CHO(g) → CH_{4}(g) + CO(g) Rate = k[CH_{3}CHO]^{3/2}

(iv) C_{2}H_{5}Cl (g) → C_{2}H_{4}(g) + HCl (g) Rate = k[C_{2}H_{5}Cl]

Order is sum of all powers of the concentrations of the reactant in “rate law expression”

**Answer:**

(i) 3NO(g) → N_{2}O(g)

Rate = k[NO]^{2}

Here rate law is dependent on concentration of NO only.And power raised to concentration of NO is 2. Hence __order of reaction is 2.__

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in . So, unit of K will be

i.e. [concentration]^{-1}[ time]^{-1} . If concentration is in mol L^{-1} and time is in sec. then __dimensions of k will be__ : __mol ^{-1} L sec^{-1}__

(ii) H_{2}O_{2}(aq) + 3I^{-}(aq) + 2H^{+} → 2 H_{2}O(l) + I_{3}^{-}

Rate = k[H_{2}O_{2}][I^{-}]

Here rate law is dependent on concentration of H_{2}O_{2} and I^{-}. And power raised to concentration of H_{2}O_{2} is 1, and power raised to concentration of I^{-} is 1. order is sum of powers hence, __order of given reaction is 2.__

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in [concentration/time]. So, unit of K will be [Concentration]/[time][concentration]^{2} i.e. [concentration]^{-1}[ time]^{-1}. If concentration is in mol L^{-1} and time is in sec then __dimensions of k will be____mol ^{-1} L sec^{-1}__

(iii) CH_{3}CHO(g) → CH_{4}(g) + CO(g)

Rate = k[CH_{3}CHO]^{3/2}

Here rate law is dependent on concentration of CH_{3}CHO.And power raised to CH_{3}CHO is 3/2. Hence order of reaction is 3/2.

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in [concentration/time]. So, unit of K will be [Concentration]/[time][concentration]^{3/2} i.e. [concentration]^{-1/2}[time]^{-1}.

If concentration is in mol L^{-1} and time is in sec. Then __dimensions of K will be mol ^{-1/2}L^{1/2} sec^{-1}__

(iv) C_{2}H_{5}Cl (g) → C_{2}H_{4}(g) + HCl(g)

Rate = k[C_{2}H_{5}Cl]

Here rate law is dependent on concentration of C_{2}H_{5}Cl. And power raised to C_{2}H_{5}Cl is 1. Hence order of reaction is 1.

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in [concentration/time]. So, unit of K will be [Concentration]/[time][concentration]^{1}. i.e. [time]^{-1}

If concentration is in mol L^{-1} and time is in sec. Then __dimensions of K will be sec ^{-1}.__

**Question 2.**

For the reaction :

2A + B → A_{2}B

the Rate = k[A][B]^{2} with k = 2.0 × 10^{-6} mol^{-2} L^{2} sec^{-1}. Calculate the initial rate of the reaction when [A] = 0.1 mol L^{-1} and [B] = 0.2 mol L^{-1}. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^{-1}.

**Answer:**

a) Rate = k[A][B]^{2}, Rate = 2.0 × 10^{-6}[0.1][0.2]^{2}

Rate = 8 × 10^{-9} mol L^{-1}sec^{-1}

b)

Situation when A is remained 0.06 mol L^{-1}

Now, According to rate law, Rate = k[A][B]^{2}

Rate = 2 × 10^{-6}[0.06][0.18]^{2}

i.e. Rate = 3.888 × 10^{-9} mol L^{-1}sec^{-1}

__Initial rate of reaction is 8 × 10 ^{-9} mol L^{-1}sec^{-1}. and rate when concentration of A is 0.06 mol L^{-1} is 3.888 × 10^{-9} mol L^{-1}sec^{-1}.__

**Question 3.**

The decomposition of NH_{3} on platinum surface is zero order reaction.

What are the rates of production of N_{2} and H_{2}. if k = 2.5 × 10^{-4} mol L^{-1} sec^{-1}?

**Answer:**

2NH_{3}(g) → N_{2}(g) + 3H_{2}(g)

Rate of zero order reaction is equal to rate constant. i.e. Rate = 2.5 × 10^{-4}mol L^{-1}sec^{-1}.

According to rate law,

2.5 × 10^{-4}mol L^{-1}sec^{-1} =

i.e. the rate of production of N_{2} is 2.5 × 10^{-4}mol L^{-1} sec^{-1}.

According to rate law,

i.e. rate of formation of H_{2} is 3 times rate of reaction = 3 × 2.5 × 10^{-4}mol L^{-1}sec^{-1}

= 7.5 × 10^{-4}mol L^{-1}sec^{-1}

__Rate of formation of N _{2} and H_{2} is 2.5 × 10^{-4} mol L^{-1}sec^{-1} and 7.5 × 10^{-4} mol L^{-1}sec^{-1} respectively.__

**Question 4.**

The decomposition of dimethyl ether leads to formation of CH_{4}, H_{2} and CO and the reaction rate is given by,

Rate = k[CH_{3}OCH_{3}]^{3/2}. The rate of reaction is followed by an increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether i.e. Rate = k[pCH_{3}OCH_{3}]^{3/2}. If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

**Answer:**

Rate of given chemical reaction will be represented as

.

Hence units of rate is bar min^{-1}

To find units of K, K = rate/[pCH_{3}OCH_{3}]^{3/2}

The unit of k = bar ^{-1/2}min^{-1}.

__Units of Rate- bar min ^{-1}. and Units of Rate constant K : bar ^{-1/2}min ^{-1}__

**Question 5.**

Mention the factors that affect the rate of chemical reaction?

**Answer:**

Factors affecting rate of reaction-

1) Temperature (increasing temperature increases rate of reaction)

2) Concentration or pressure of reactants. (Increasing concentration or pressure increases rate of reaction)

3) Presence or absence of a catalyst. (Adding catalyst mostly increases the rate of reaction)

4) The surface area of solid reactant. (If reaction is processing over solid reactant then increasing its surface area increases rate of reaction)

5) Nature of reactants

**Question 6.**

A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration is

(i) doubled (ii) reduced to half

**Answer:**

Let suppose reaction

A ⇒ B;

having rate law, Rate = K [A]^{2}

(i) If the concentration of A is doubled then the rate will affect by the square of concentration i.e. rate will become 2^{2} = 4 times.

(ii) If the concentration of A is halved then the rate will affect the square of concentration i.e (1/2)^{2} = 1/4 times.

__(i) Rate becomes 4 times of initial Rate (ii) Rate becomes 1/4 times of initial Rate__

**Question 7.**

What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?

**Answer:**

__Increase in temperature increases the rate constant of a reaction__. as we know increase in temperature increases the rate of reaction to satisfy the equation Rate = k [concentration]^{n} where n can be any real number. k have to increase as concentration is almost not changing over small temperature change. Increasing temperature by 10^{0}C almost double the rate constant.

This can be represented quantitively by the help of arrehinus equation-

__K = Ae ^{-Ea/RT}__, where k is rate constant, E

_{a}is activation energy, R is universal gas constant, T is absolute temperature.

**Question 8.**

In a pseudo first order hydrolysis of ester in water, the following results were obtained

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds

(ii) Calculate the pseudo first-order rate constant for the hydrolysis of the ester.

**Answer:**

(i) Average rate of reaction over interval is [change in concentration]/[time taken] i.e.

(ii) the pseudo first-order rate constant can be calculated by

K = (2.303/t) log(C_{i}/C_{t}) where K is Rate constant,

t is time taken,

C_{i} is initial concentration

C_{t} is Concentration at time t.

K = (2.303/30) log (0.55/0.31)

⇒ K = 1.9 × 10^{-2} sec^{-1}

__(i) Average rate between 30 to 60 sec is 0.00467 mol L ^{-1}sec^{-1}__

__(ii) Pseudo first order rate constant is 1. × 10 ^{-2}sec^{-1}__

**Question 9.**

A reaction is First Order in A and second order in B.

(i) Write the differential rate equation.

(ii) How is the rate affected on increasing the concentration of B three times?

(iii) How is the rate affected when the concentrations of both A and B are doubled?

**Answer:**

(i) Order is power raised to reactant in rate law, hence,

Rate = k[A][B]^{2}

(ii) When the concentration of B is increased three times then the rate is affected by the square of reactant. The rate is increased by 9 Times.

(iii) When concentration of reactant both A and B is doubled then the rate will have affected as square of reactant B and Two times of Reactant A. Overall increase in rate is 8 times

__(i) When the concentration of B is increased by three times, the rate is increased by nine times.__

__(ii) When conc. of both reactants doubled then Rate increased 8 times.__

**Question 10.**

In a reaction between A and B, the initial rate of reaction(r_{0})was measured for different initial concentrations of A and B are given below:

What is the order of reaction with respect to A and B?

**Answer:**

When concentration of B is changed then rate of reaction doesn't change that means order with respect to B is 0. But when the concentration of A is doubled rate increased by 2.82 times i.e.2^{1.5} = 2.82.Hence order with respect to A is 1.5.

Order with respect to A and B is 1.5 and 0 respectively.

**Question 11.**

The following results have been obtained during the kinetic studies of the reaction:

2A + B → C + D

Determine the Rate law and the rate constant for the reaction.

**Answer:**

By comparing Experiment I and IV if we increase the concentration of A by 4 times then Rate also increased by 4 times. That means order with respect to A is 1.

By comparing Experiment II and III if we double the concentration of B Rate increases by 4 times that means order with respect to B is 2.

Rate law of reaction will be, Rate = k [A][B]^{2}

To find K, K = rate/[A][B]^{2} i.e. K = 6.0 × 10^{-3}/[0.1][0.1]^{2}

K = 6 mol^{-2}L^{2}sec^{-1}

__Order with respect to A and B is 1 and 2 Respectively. And value of K(rate constant) is = 6 mol ^{-2}L^{2}sec^{-1}__

**Question 12.**

The reaction between A and B is first order with respect to A and zero order with respect to B.Phil. in the blanks in the following table:

**Answer:**

As reaction is first order with respect to A and zero Order with respect to B. Then changing the concentration of B won’t affect the rate of reaction and increasing concentration of A ‘n’ times will increase the rate by ‘n’ times. By this logic lets fill the table- In first blank space concentration of A will be 0.2 mol L^{-1} because the rate is doubled. In second blank space, Rate will be 8 × 10^{-2}mol L^{-1}min^{-1} because the concentration of A is increased 4 Times. In third blank space concentration of A will be 0.1 mol L^{-1} because the rate is same as in experiment I.

**Question 13.**

Calculate the half life of a first order reaction from their rate constants given below:

(i) 200 s^{-1} (ii) 2 min^{-1} (iii) 4 years^{-1}

**Answer:**

Half life of first order reaction is, t_{1/2} = ln2/K where t_{1/2} is half life of first order reaction, K is rate constant of First order reaction.

(i) t_{1/2} = ln2/200 s^{-1}

⇒ t_{1/2} = 0.693/200 s^{-1} (∵ ln2 = 0.693)

⇒ t_{1/2} = 0.003465 sec.

(ii) t_{1/2} = ln2/2 min^{-1}

⇒ t_{1/2} = 0.693/2 min^{-1} (∵ ln2 = 0.693)

⇒ t_{1/2} = 0.3465 min

(iii) t_{1/2} = ln2/4 year^{-1}

⇒ t_{1/2} = 0.693/4 years ^{-1} (∵ ln2 = 0.693)

⇒ t_{1/2} = 0.17325 year.

__Half life of 3 reactions are 0.003465 sec, 0.3465 min, 0.17325 year respectively.__

**Question 14.**

The half-life for radioactive decay of 14C is 5730 years. An archeological artifact containing wood had only 80%of the 14C found in a living tree. Estimate the age of the sample.

**Answer:**

Radio active decay occurs via first order rate law,

t_{1/2} = 5730 years. rate constant(k) of given decay is 0.693/t_{1/2}

⇒ 0.693/5730 = 1.2 × 10^{-4} year^{-1}

By first order integrated rate law age of sample will be,

where T is the age of sample A_{0} is the initial activity of the sample. and A_{t} is the activity of sample at any time t.

T = 0.18 × 10^{4} years.

__Age of given sample is 0.18 × 10 ^{4} years.__

**Question 15.**

The experiemental data for decomposition of N_{2}O_{5}

[2N_{2}O_{5}→ 4NO_{2} + O_{2}]

in gas phase at 318 k are given below:

(i) plot N_{2}O_{5} against t

(ii) Find the half-life period for the Reaction.

(iii) Draw a graph between log[N_{2}O_{5}]and t.

(iv) what is the rate law?

(v) Calculate the rate constant.

(vi) Calculate the half-life period from k and compare it with (ii)

**Answer:**

(i)

(ii) initial conc.is 1.63 × 10^{-2} so half conc. is 0.815 × 10^{-2}M From given information and graph t_{1/2} is 1440 sec.

(iii)

(iv) As log [N_{2}O_{5}] vs time is straight line given reaction is first order. Hence its rate law will be, Rate = k[N_{2}O_{5}]

(v) The slope of above graph is slope = 0.000209

K = 2.303 × slope

⇒ 4.82 × 10^{-4}sec^{-1}

Now, t_{1/2} = 0.693/K.

⇒ 0.693/4.82 × 10^{-4}

⇒ t_{1/2} = 1438 sec. which is almost equal to (ii)

**Question 16.**

The rate constant for a first order reaction is 60 s^{–1}. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

**Answer:**

Given:

Order of the reaction = 1

Let, Initial concentration [R]^{°}= x

Final concentration [R] = x/16

Rate constant k = 60 s^{-1}

We know, time

⇒

⇒

Solving, we get __t = 4.6 × 10 ^{-2}s__

**Question 17.**

During nuclear explosion, one of the products is ^{90}Sr with half-life of 28.1 years. If 1Î¼g of ^{90}Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

**Answer:**

Initial concentration, [R]_{°} = 1Î¼g

Final concentration, [R]

Half-life t_{1/2} = 28.1 yrs

Solution:

We know, t_{1/2} = 0.693/k

Where, k – rate constant

⇒ k = 0.693/ t_{1/2}

⇒ k = 0.693/(28.1 yrs)

⇒ k = 0.0246 yrs^{-1}

Also,

If t = 10yrs, then, using the formula, we get,

⇒

⇒

⇒

⇒

⇒

⇒

⇒ = 0.7824Î¼g

If t = 60yrs, then again, we get,

⇒

⇒

⇒

⇒

⇒ [R]_{60} = 0.228 Î¼g

∴ 0.2278Î¼g of ^{90}Sr will be left after 60 years.

**Question 18.**

For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

**Answer:**

Let, initial concentration be [R]_{°}

Concentration at 90% completion be ((100-90)/100)×[R]_{°}

∴ Concentration at 90% be 0.1[R]_{°}

Concentration at 99% completion be ((100-99)/100)× [R]_{°}∴ Concentration at 99% be 0.01[R]_{°}

We know, time

Time taken for 90% completion is

⇒

⇒

Time taken for 99% completion is

⇒

⇒

⇒ t_{99} = 2t_{90}

Hence, the time taken to complete 9% of the first order reaction is twice the time required for the completion of 90% of the reaction.

**Question 19.**

A first order reaction takes 40 min for 30% decomposition. Calculate t_{1/2}.

**Answer:**

Given:

Time t = 40 min

When 30% decomposition is undergone, 70% is the concentration.

We know, time taken

Where, k- rate constant

[R]_{°} -Initial concentration

[R]-Concentration at time ‘t’

⇒

⇒

⇒

⇒ 258.23 = (2.303/k)

We know, Half-life t_{1/2} = 0.693/k

Which can be written as, t_{1/2} = 0.3010 × (2.303/k)

⇒ t_{1/2} = 0.3010 × 258.23

⇒ t_{1/2} = 77.72 min

**Question 20.**

For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

Calculate the rate constant.

**Answer:**

When t = 0, the total partial pressure is P_{0} = 35.0 mm of Hg

When time t = t, the total partial pressure is P_{t} = P_{0} + p

P_{0}-p = P_{t}-2p, but by the above equation, we know p = P_{t}-P_{0}

Hence, P_{0}-p = P_{t}-2( P_{t}-P_{0})

Thus, P_{0}-p = 2P_{0} – P_{t}

We know that time

Where, k- rate constant

[R]_{°} -Initial concentration of reactant

[R]-Concentration of reactant at time ‘t’

Here concentration can be replaced by the corresponding partial pressures.

Hence, the equation becomes,

→ equation 1

At time t = 360 s, P_{t} = 54 mm of Hg and P_{0} = 30 mm of Hg,

Substituting in equation 1,

⇒

⇒ k = 2.175 × 10^{-3} s^{-1}

At time t = 720 s, P_{t} = 63 mm of Hg and P_{0} = 30 mm of Hg,

Substituting in equation 1,

⇒

Thus, k = 2.235 × 10^{-3} s^{–1}

Taking average, k = (2.235 × 10^{-3} s^{–1} + 2.175 × 10^{-3} s^{–1})/2

∴ k = 2.21 × 10^{-3} s^{–1}.

**Question 21.**

The following data were obtained during the first order thermal decomposition of SO_{2}Cl_{2} at a constant volume.

SO_{2}Cl_{2}_{(g)}→ SO_{2 (g)} + Cl_{2 (g)}

Calculate the rate of the reaction when total pressure is 0.65 atm

**Answer:**

When t = 0, the total partial pressure is P_{0} = 0.5 atm

When time t = t, the total partial pressure is P_{t} = P_{0} + p

P_{0}-p = P_{t}-2p, but by the above equation, we know p = P_{t}-P_{0}

Hence, P_{0}-p = P_{t}-2(P_{t}-P_{0})

Thus, P_{0}-p = 2P_{0} – P_{t}

We know that time

Where, k- rate constant

[R]_{°} -Initial concentration of reactant

[R]-Concentration of reactant at time ‘t’

Here concentration can be replaced by the corresponding partial pressures.

Hence, the equation becomes,

⇒ → equation 1

At time t = 100 s, P_{t} = 0.6 atm and P_{0} = 0.5 atm,

Substituting in equation 1,

⇒

Thus, k = 2.231 × 10^{-3} s^{-1}

The rate of reaction R = k × P_{S02Cl2}

When total pressure P_{t} = 0.65 atm and P_{0} = 0.5 atm, then

P_{S02Cl2} = 2P_{0}-P_{t}

Thus, substituting the values, P_{S02Cl2} = 2(0.5)-0.6 = 0.35 atm

R = k × P_{S02Cl2} = 2.231 × 10^{-3} s^{–1} × 0.35

Rate of the reaction R = 7.8 × 10^{-4}atm s^{–1}

**Question 22.**

The rate constant for the decomposition of N_{2}O_{5} at various temperatures is given below:

Draw a graph between ln k and 1/T and calculate the values of A and Ea. Predict the rate constant at 30° and 50°C.

**Answer:**

To convert the temperature in °C to °K we add 273 K.

The graph is given as:

The Arrhenius equation is given by k = Ae^{-Ea/RT}

Where, k- Rate constant

A- Constant

E_{a}-Activation Energy

R- Gas constant

T-Temperature

Taking natural log on both sides,

ln k = ln A-(E_{a}/RT)

→ equation 1

By plotting a graph, ln K Vs 1/T, we get y-intercept as ln A and Slope is –E_{a}/R.

Slope = (y_{2}-y_{1})/(x_{2}-x_{1})

By substituting the values, slope = -12.301

⇒ –E_{a}/R = -12.301

But, R = 8.314 JK^{-1}mol^{-1}

⇒ E_{a} = 8.314 JK^{-1}mol^{-1} × 12.301 K

⇒ E_{a} = 102.27 kJ mol^{-1}

Substituting the values in equation 1 for data at T = 273K

⇒

⇒

(∵ At T = 273K, ln k = -7.147)

On solving, we get ln A = 37.911

∴ A = 2.91×10^{6}

When T = 30^{0}C, hence T = 30 + 273 = 303K

⇒

⇒

⇒ ln k = -2.8

⇒ k = 6.08x10^{-2}s^{-1}.

When T = 50^{0}C, hence T = 50 + 273 = 323K

Substituting in equation 1,

A = 2.91 × 10^{6}, E_{a} = 102.27 kJ mol^{-1}

Ln K = ln(2.91 × 10^{6})- 102.27/(8.314 × 323)

Thus, ln k = -0.5 and k = 0.607s^{-1}.

**Question 23.**

The rate constant for the decomposition of hydrocarbons is 2.418 × 10^{–5}s^{–1} at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

**Answer:**

Given,

k = 2.418 × 10^{-5} s^{-1}

T = 546 K

E_{a} = 179.9 kJ mol^{-1} = 179.9 × 10^{3}J mol^{-1}

The Arrhenius equation is given by k = Ae^{-Ea/RT}

Taking natural log on both sides,

Ln k = ln A-(E_{a}/RT)

Substituting the values,

ln(2.418 × 10^{-5} ) = ln A-179.9/(8.314 × 546)

ln A = 12.5917

A = 3.9 × 10^{12} s^{-1}(approximately)

**Question 24.**

Consider a certain reaction A → Products with k = 2.0 × 10^{–}^{2}s^{-}^{1}. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L^{–1}.

**Answer:**

Given,

k = 2.0 × 10^{–}^{2}s^{-}^{1}

time t = 100s

Concentration [A_{0}] = 1.0 mol L^{-1}

We know,

On substituting the values,

Log(1/[A]) = 2.303/2

Log[A] = -2.303/2

[A] = 0.135 mol L^{–1}

**Question 25.**

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_{1/2} = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

**Answer:**

t_{1/2} = 3.00 hours

We know, t_{1/2} = 0.693/k

∴ k = 0.693/3

k = 0.231hrs^{-1}

We know, time

Where, k- rate constant

[R]_{°} -Initial concentration

[R]-Concentration at time ‘t’

Thus, substituting the values,

log([R]_{0}/[R]) = 0.8

log([R]/[R]_{0}) = -0.8

[R]/[R]_{0} = 0.158

Hence, 0.158 fraction of sucrose remains.

**Question 26.**

The decomposition of hydrocarbon follows the equation

k = (4.5 × 10^{11}s^{–1}) e^{-28000K}/T. Calculate E_{a}.

**Answer:**

The given equation is

k = (4.5 x 10^{11} s^{-1}) e^{-28000 K/T} (i)

Comparing, Arrhenius equation

k = Ae ^{-E}_{a}^{/RT} (ii)

We get, E_{a} / RT = 28000K / T

⇒ E_{a} = R x 28000K

= 8.314 J K^{-1}mol^{-1} × 28000 K

= 232792 J mol^{–1} or 232.792 kJ mol^{–1}

**Question 27.**

The rate constant for the first order decomposition of H_{2}O_{2} is given by the following equation: log k = 14.34 – 1.25 × 10^{4}K/T

Calculate E_{a} for this reaction and at what temperature will its half-period be 256 minutes?

**Answer:**

We know, The Arrhenius equation is given by k = Ae^{-Ea/RT}

Taking natural log on both sides,

Ln k = ln A-(E_{a}/RT)

Thus, log k = log A -(E_{a}/2.303RT) → eqn 1

The given equation is log k = 14.34 – 1.25 × 10^{4}K/T → eqn 2

Comparing 2 equations,

E_{a}/2.303R = 1.25 × 10^{4}K

E_{a} = 1.25 × 10^{4}K × 2.303 × 8.314

E_{a} = 239339.3 J mol^{-1} (approximately)

E_{a} = 239.34 kJ mol^{-1}

Also, when t_{1/2} = 256 minutes,

k = 0.693 / t_{1/2}

= 0.693 / 256

= 2.707 × 10^{-3} min^{-1}

k = 4.51 × 10^{-5}s^{–1}

Substitute k = 4.51 × 10^{-5}s^{–1} in eqn 2,

log 4.51 × 10^{-5} s^{–1} = 14.34 – 1.25 × 10^{4}K/T

log(0.654-5) = 14.34– 1.25 × 10^{4}K/T

T = 1.25 × 10^{4}/[ 14.34- log(0.654-5)]

T = 668.9K or T = 669 K

**Question 28.**

The decomposition of A into product has value of k as 4.5 × 10^{3} s^{–1} at 10°C and energy of activation 60 kJ mol^{–1}. At what temperature would k be 1.5 × 10^{4}s^{–1}?

**Answer:**

From Arrhenius equation, we obtain

Also, k_{1} = 4.5 × 10^{3} s ^{- 1}

T_{1} = 273 + 10 = 283 K

k_{2} = 1.5 × 10^{4} s ^{- 1}

E_{a} = 60 kJ mol ^{- 1} = 6.0 × 10^{4} J mol ^{- 1}

Then,

→ 0.5229 = 3133.627 × (T_{2}-283)/(283 × T_{2})

→ 0.0472T_{2} = T_{2}-283

T_{2} = 297K or T_{2} = 24^{0} C

**Question 29.**

The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 10^{10}s^{–1}. Calculate k at 318K and E_{a}.

**Answer:**

We know, time t = (2.303/k) × log([R]_{0}/[R])

Where, k- rate constant

[R]_{°} -Initial concentration

[R]-Concentration at time ‘t’

At 298K, If 10% is completed, then 90% is remaining.

t = (2.303/k) × log ([R]_{0}/0.9[R]_{0})

t = (2.303/k) × log (1/0.9)

t = 0.1054 / k

At temperature 308K, 25% is completed, 75% is remaining

t’ = (2.303/k’) × log ([R]_{0}/0.75[R]_{0})

t’ = (2.303/k’) × log (1/0.75)

t’ = 2.2877 / k'

But, t = t’

0.1054 / k = 2.2877 / k'

k' / k = 2.7296

From Arrhenius equation, we obtain

log k_{2}/k_{1} = (E_{a} / 2.303 R) × (T_{2} - T_{1}) / T_{1}T_{2}

Substituting the values,

⇒

E_{a} = 76640.09 J mol^{-1} or 76.64 kJ mol^{-1}

We know, log k = log A –E_{a}/RT

Log k = log(4 × 10^{10})-(76.64kJ mol^{-1}/(8.314 × 318))

Log k = -1.986

∴ k = 1.034 x 10^{-2} s ^{-1}

**Question 30.**

The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

**Answer:**

Given, k_{2} = 4k_{1}, T_{1} = 293K and T_{2} = 313K

We know, From Arrhenius equation, we obtain

⇒

⇒

On solving we get,

E_{a} = 58263.33 J mol^{-1} or 58.26 kJ mol^{-1}