Chapter 16, Problem 81A

Interpretation Introduction

Interpretation:

To calculate the reaction’s initial, instantaneous rate of the reaction ${\text{NO}}_{\text{2}}\left(\text{g}\right)+\text{CO}\left(\text{g}\right)\to \text{NO}\left(\text{g}\right){\text{+CO}}_{\text{2}}\left(\text{g}\right)$.

Concept introduction:

Consider a general reaction

$\text{aA+bB}\to \text{Products}$

The rate law of the above reaction is

$\text{rate}=k{\left[\text{A}\right]}^m{\left[\text{B}\right]}^n$

Here, $\left[\text{A}\right]$ and $\left[\text{B}\right]$ is the concentration of reactant A and B respectively, k represents specific rate and m and n represents order of corresponding reactants.

Answer to Problem 81A

Instantaneous rate of the reaction is $\text{1}\text{.2}\times {\text{10}}^{-5}\text{L/mol}\text{.s}$.

Explanation of Solution

The reaction is as follows:

${\text{NO}}_{\text{2}}\left(\text{g}\right)+\text{CO}\left(\text{g}\right)\to \text{NO}\left(\text{g}\right){\text{+CO}}_{\text{2}}\left(\text{g}\right)$

The reaction rate at particular temperature is given below.

$\text{Rate}=0.50\text{L/mol}\text{.s}{\left[{\text{NO}}_{\text{2}}\right]}^2$ …… (1)

Given,

$\left[{\text{NO}}_{\text{2}}\right]=0.0048\text{mol/L}$

Substitute the above values in equation (1).

$\begin{array}{c}\text{Rate}=0.50\text{L/mol}\text{.s}{\left[{\text{NO}}_{\text{2}}\right]}^2\\ =0.50\text{L/mol}\text{.s}{\left[0.0048\text{}\text{mol/L}\right]}^2\\ =0.00002304\times 0.50\text{L/mol}\text{.s}\\ \text{=0}\text{.00001152}\text{L/mol}\text{.s}\end{array}$

$\text{rate}\text{=1}\text{.2}\times {\text{10}}^{-5}\text{L/mol}\text{.s}$

Therefore, rate of the reaction is $\text{1}\text{.2}\times {\text{10}}^{-5}\text{L/mol}\text{.s}$.

Conclusion

Instantaneous rate of the reaction is $\text{1}\text{.2}\times {\text{10}}^{-5}\text{L/mol}\text{.s}$.