Question

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided. (4 points)
A + B yieldsproducts

Trial

[A]

[B]

Rate

1 0.10 M 0.20 M 1.2 × 10-2 M/min
2 0.10 M 0.40 M 4.8 × 10-2 M/min
3 0.20 M 0.40 M 9.6 × 10-2 M/min

Answer 1

Answer: Rate law=$k[A]^1[B]^2$, order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is $3L^2mol^{-2}s^{-1}$

Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

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

k= rate constant

x = order with respect to A

y = order with respect to A

n = x+y = Total order

a) From trial 1: $1.2\times 10^{-2}=k[0.10]^x[0.20]^y$    (1)

From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$    (2)

Dividing 2 by 1 :$\frac{4.8\times 10^{-2}}{1.2\times 10^{-2}}=\frac{k[0.10]^x[0.40]^y}{k[0.10]^x[0.20]^y}$

$4=2^y,2^2=2^y$ therefore y=2.

b) From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$    (3)

From trial 3: $9.6\times 10^{-2}=k[0.20]^x[0.40]^y$   (4)

Dividing 4 by 3:$\frac{9.6\times 10^{-2}}{4.8\times 10^{-2}}=\frac{k[0.20]^x[0.40]^y}{k[0.10]^x[0.40]^y}$

$2=2^x,2=2^1$, x=1

Thus rate law is $Rate=k[A]^1[B]^2$

Thus order with respect to A is 1 , order with respect to B is 2 and total order is 1+2=3.

c) For calculating k:

Using trial 1:  $1.2\times 10^{-2}=k[0.10]^1[0.20]^2$

$k=3 L^2mol^{-2}s^{-1}$.