Chapter 16.4, Problem 32PP

Interpretation Introduction

Interpretation:

The instantaneous rate for the given reaction needs to be determined.

${\text{2 NO}}_{\text{(g) }}{\text{+ H}}_{\text{2(g)}}\stackrel{}{\to }{\text{N}}_{\text{2}}{\text{O}}_{\text{(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

Concept introduction:

Chemical kinetic is the branch of chemistry that deals with kinetic or rate of a chemical reaction. The rate of reaction can be defined as the change in the concentration of reactant and product with time. The rate of reaction mainly depends on the activation energy.

Answer to Problem 32PP

$\text{Rate = }3.62{\text{ × 10}}^{\text{-5}}\text{ M/sec}$

Explanation of Solution

Given information:

$\begin{array}{l}\text{Rate constant = 2}{\text{.90 × 10}}^{\text{2}}{\text{ L}}^{\text{2}}{\text{/mol}}^{\text{2}}\text{.s}\\ \text{[NO] = 0}\text{.0100 M}\\ {\text{[H}}_{\text{2(g)}}\text{]=0}\text{.00125M}\end{array}$

The instantaneous rate of reaction represents the change in concentration of reactant at an instant time. It can be written as:

$\text{instantaneous}\text{rate = }\frac{\text{Δ[x]}}{\text{Δt}}$

The rate law must be written as:

${\text{Rate = k [NO]}}^{\text{2}}{\text{ [H}}_{\text{2}}\text{] }$

Substitute the values to calculate the rate:

$\begin{array}{l}{\text{Rate = k [NO]}}^{\text{2}}{\text{ [H}}_{\text{2}}\text{] }\\ \text{Rate = 2}{\text{.90 × 10}}^{\text{2}}{\text{ L}}^{\text{2}}{\text{/mol}}^{\text{2}}\text{.s [0}{\text{.0100 M]}}^{\text{2}}\text{ [0}\text{.00125 M]}\\ \text{Rate = }3.62{\text{ × 10}}^{\text{-5}}\text{ M/sec}\end{array}$

Conclusion

Thus,

$\text{Rate = }3.62{\text{ × 10}}^{\text{-5}}\text{ M/sec}$