Question

Determine the rate law and the value of k for the following reaction using the data provided.2 NO(g) + O2(g) → 2 NO2(g)[NO]i (M)[O2]i (M)Initial Rate (M-2s-1)0.0300.00558.55 x 10-30.0300.01101.71 x 10-20.0600.00553.42 x 10-2

Answer 1

Answer: Rate law=$Rate=k[NO]^2[O_2]^1$

Rate law constant is $1727.3L^2mol^{-2}s^{-1}$

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.

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

$Rate=k[NO]^x[O_2]^y$

k= rate constant

x = order with respect to NO

y = order with respect to $O_2$

n = x+y = Total order

a) From trial 1: $8.55 x 10^{-3}=k[0.030]^x[0.0055]^y$    (1)

From trial 2: $1.71 x 10^{-2}=k[0.030]^x[0.0110]^y$    (2)

Dividing 2 by 1 :$\frac{1.71\times 10^{-2}}{8.55\times 10^{-3}}=\frac{k[0.030]^x[0.0110]^y}{k[0.030]^x[0.0055]^y}$

$2=2^y,2^1=2^y$ therefore y=1.

b) From trial 1 :$8.55 x 10^{-3}=k[0.030]^x[0.0055]^y$   (3)

From trial 3:$3.42\times 10^{-2}=k[0.060]^x[0.0055]^y$ (4)

Dividing 4 by 3:$\frac{3.42\times 10^{-2}}{8.55\times 10^{-3}}=\frac{k[0.060]^x[0.0055]^y}{k[0.030]^x[0.0055]^y}$

$4=2^x,2^2=2^x$, x=2Thus rate law is $Rate=k[NO]^2[O_2]^1$

Thus order with respect to $NO$ is 2 , order with respect to $O_2$ is 1 and total order is 1+2=3.

Rate law is $Rate=k[NO]^2[O_2]^1$

b) For calculating k:

Using trial 1:  $8.55\times 10^{-3}=k[0.030]^2[0.0055]^1$

$k=1727.3L^2mol^{-2}s^{-1}$

The value of rate constant is $1727.3L^2mol^{-2}s^{-1}$