Question

Consider this initial-rate data at a certain temperature for the reaction described by

2NO2(g)+O3(g) ----> N2O5(g) + O2(g)

[NO2]0 (M)         [O3]0(M)      Initial rate (M/s)

0.650                  0.800           2.47x10^4

1.10                    0.800           4.18x10^4

1.76                    1.40             11.70x10^4

Determine the value and units of the rate constant.

K=_____

Answer 1

Concepts and reason

This problem is based on the concept of chemical kinetics.

The rate constant of a reaction is determined by rate of the reaction. For this, first order of the reaction is calculated.

Fundamentals

For the reaction,

${\rm{A}} + {\rm{B}} \to {\rm{C}} + {\rm{D}}$

The rate of the reaction is written as follows:

$r = k{\left[ A \right]^x}{\left[ B \right]^y}$

Here, $\left[ A \right]$ , $\left[ B \right]$ are the molar concentration of reactants A and B respectively. And x and y are order of the reaction with respect to A and B.

The reaction is as follows:

${\rm{2N}}{{\rm{O}}_2}\left( g \right) + {{\rm{O}}_3}\left( g \right) \to {{\rm{N}}_2}{{\rm{O}}_5}\left( g \right) + {{\rm{O}}_2}\left( g \right)$

The rate of the reaction is written as follows:

$r = k{\left[ {{\rm{N}}{{\rm{O}}_2}} \right]^x}{\left[ {{{\rm{O}}_3}} \right]^y}$

The order of reaction is calculated by dividing the two rate law equations as follows:

$\frac{{2.47 \times {{10}^4}}}{{4.18 \times {{10}^4}}} = \frac{{k{{\left[ {{\rm{N}}{{\rm{O}}_2}} \right]}^x}{{\left[ {{{\rm{O}}_3}} \right]}^y}}}{{k{{\left[ {{\rm{N}}{{\rm{O}}_2}} \right]}^x}{{\left[ {{{\rm{O}}_3}} \right]}^y}}}$

Here, x is order with respect to concentration of ${\rm{N}}{{\rm{O}}_2}$ , y is order with respect to concentration of ${{\rm{O}}_3}$ .

Now, according to the table, the order of reaction is calculated as follows:

$\begin{array}{c}\\\frac{{2.47 \times {{10}^4}}}{{4.18 \times {{10}^4}}} = \frac{{k{{\left[ {0.65} \right]}^x}{{\left[ {0.8} \right]}^y}}}{{k{{\left[ {1.1} \right]}^x}{{\left[ {0.8} \right]}^y}}}\\\\0.59 = {\left[ {0.59} \right]^x}\\\\x = 1\\\end{array}$

Thus, the order of reaction with respect to ${\rm{N}}{{\rm{O}}_2}$ is 1.

Also, Now divide the equations as follows:

$\begin{array}{c}\\\frac{{2.47 \times {{10}^4}}}{{11.7 \times {{10}^4}}} = \frac{{k{{\left[ {0.65} \right]}^1}{{\left[ {0.8} \right]}^y}}}{{k{{\left[ {1.76} \right]}^1}{{\left[ {1.4} \right]}^y}}}\\\\{\left[ {0.571} \right]^1} = {\left[ {0.571} \right]^y}\\\\y = 1\\\end{array}$

Thus, the order of reaction with respect to ${{\rm{O}}_3}$ is 1.

Thus, rate law equation is written as follows:

$r = k{\left[ {{\rm{N}}{{\rm{O}}_2}} \right]^1}{\left[ {{{\rm{O}}_3}} \right]^1}$

Rearrange for the rate constant as follows:

$k = \frac{r}{{{{\left[ {{\rm{N}}{{\rm{O}}_2}} \right]}^1}{{\left[ {{{\rm{O}}_3}} \right]}^1}}}$

Now, substitute $11.7 \times {10^4}$ for r , 1.76 for $\left[ {{\rm{N}}{{\rm{O}}_2}} \right]$ and 1.40 for $\left[ {{{\rm{O}}_3}} \right]$ .

$\begin{array}{c}\\k = \frac{{11.7 \times {{10}^4}}}{{\left( {1.76} \right)\left( {1.40} \right)}}\\\\ = 4.74 \times {10^4}{\rm{ mo}}{{\rm{l}}^{ - 1}}{\rm{ L se}}{{\rm{c}}^{ - 1}}\\\end{array}$

Ans:

The rate constant is $4.74 \times {10^4}{\rm{ mo}}{{\rm{l}}^{ - 1}}{\rm{ L se}}{{\rm{c}}^{ - 1}}$ .