# The following results have been obtained during the kinetic studies of the reaction: 2A + B → C + D - Chemistry

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

2A + B → C + D

 Experiment A/ mol L−1 B/ mol L−1 Initial rate of formation of D/mol L−1 min−1 I 0.1 0.1 6.0 × 10−3 II 0.3 0.2 7.2 × 10−2 III 0.3 0.4 2.40 × 10−2 IV 0.4 0.1

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

#### Solution

Let the order of the reaction with respect to A be x and with respect to B be y.

Therefore, rate of the reaction is given by,

Rate = k[A]^x[B]^y

According to the question,

6.0 xx 10^(-3) = k[0.1]^x [0.1]^y .....(i)

7.2 xx 10^(-2) = k[0.3]^x[0.2]^y....(ii)

2.88xx10^(-1) = k[0.3]^x[0.4]^y ....(iii)

2.40 xx 10^(-2) = k[0.4]^x[0.1]^y .....(iv)

Dividing equation (iv) by (i), we obtain

(2.40xx10^(-2))/(6.0xx10^(-3)) = (k[0.4]^x[0.1]^y)/(k[0.1]^x[0.1]^y)

=> 4 = [0.4]^x/[0.1]^x

=> 4 = (0.4/0.1)^x

=>(4)^1 = 4^x

=> x = 1

Dividing equation (iii) by (ii), we obtain

(2.88xx10^(-1))/(7.2xx10^(-2)) = (k[0.3]^x[0.4]^y)/(k[0.3]^x[0.2]^y)

=> 4 = (0.4/0.2)^y

=> 4 = 2y

=> 22 = 2y

=> y = 2

Therefore, the rate law is

Rate = [A] [B]2

=>k = "Rate"/[A][B]^2

From experiment I, we obtain

k = (6.0xx10^(-3) "mol L"^(-1) "min"^(-1))/((0.1 "mol L"^(-1))(0.1 "mol L"^(-1))^2)

= 6.0 L2 mol−2 min−1

From experiment III, we obtain

k = (7.2 xx10^(-2) "mol L"^(-1) "min"^(-1))/(0.3 "mol L"^(-1)(0.2 "mol L"^(-1))^2)

= 6.0 L2 mol−2 min−1

From experiment III, we obtain

k = (2.88xx10^(-1) mol L^(-1) min^(-1))/((0.3 "mol L"^(-1))(0.4 "mol L"^(-1))^2)

= 6.0 L2 mol−2 min−1

From experiment IV, we obtain

k = (2.40xx10^(-2) mol L^(-1) "min"^(-1))/(0.4 mol L^(-1)(0.1 "mol L"^(-1))^2)`

= 6.0 L2 mol−2 min−1

Therefore, rate constant, k = 6.0 L2 mol−2 min−1

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Chapter 4: Chemical Kinetics - Exercises [Page 118]

#### APPEARS IN

NCERT Chemistry Class 12
Chapter 4 Chemical Kinetics
Exercises | Q 11 | Page 118
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