Chapter 16.4, Problem 33PP

Interpretation Introduction

Interpretation:

The concentration of [NO] needs to be determined, if the rate is $9.00{\text{ × 10}}^{\text{-5}}\text{ M/sec}$ and [H₂] is 0.00300M for the given reaction.

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

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 33PP

$\text{[NO] = 1}{\text{.02× 10}}^{\text{-2}}\text{M}$

Explanation of Solution

Given information:

$\begin{array}{l}\text{Rate = }9.00{\text{ × 10}}^{\text{-5}}\text{ M/sec}\\ {\text{[H}}_{\text{2(g)}}\text{]=0}\text{.00300 M}\\ \text{Rate constant = 2}{\text{.90× 10}}^2{\text{ L}}^{\text{2}}{\text{/(mol}}^{\text{2}}\text{.s)}\end{array}$

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:

$\begin{array}{l}{\text{Rate = k [NO]}}^{\text{2}}{\text{ [H}}_{\text{2}}\text{] }\\ \text{9}{\text{.00 × 10}}^{\text{-5}}\text{ M/sec = 2}{\text{.90 × 10}}^{\text{2}}{\text{ L}}^{\text{2}}{\text{/mol}}^{\text{2}}{\text{.s× [NO]}}^{\text{2}}\text{ ×[0}\text{.00300 M]}\\ {\text{ [NO]}}^{\text{2}}\text{ = }\frac{\text{9}{\text{.00 × 10}}^{\text{-5}}\text{ M/sec }}{\text{2}{\text{.90 × 10}}^{\text{2}}{\text{ L}}^{\text{2}}{\text{/mol}}^{\text{2}}\text{.s×[0}\text{.00300 M]}}\\ {\text{[NO]}}^{\text{2}}\text{ = 1}\text{.0344}\\ \text{[NO] = 1}{\text{.02× 10}}^{\text{-2}}\text{M}\end{array}$

Conclusion

Thus,

$\text{[NO] = 1}{\text{.02× 10}}^{\text{-2}}\text{M}$