Chapter 7.4, Problem 7E

a.

To determine

To find : an expression for the concentration $\left[N_2{O}_5\right]$ after $t$ seconds if the initial concentration is $C$

Answer to Problem 7E

  $\left[N_2{O}_5\right]\left(t\right)=C\times e^{-0.0005t}$

Explanation of Solution

Given information : The question is

The chemical equation is $N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2$ takes place at ${45}^{\circ }C$ , the rate of reaction of dinitrogen pent oxide is proportional to its concentration

  $-\frac{d\left[N_2{O}_5\right]}{dt}=0.0005\left[N_2{O}_5\right]$

Calculation : The provided question can be rewritten in terms of $y$ if replace $-\frac{d\left[N_2{O}_5\right]}{dt}$ with $-\frac{dy}{dt}$ and $0.0005\left[N_2{O}_5\right]$ with $0.0005y$

  $\begin{array}{l}-\frac{dy}{dt}=0.0005y\\ \frac{dy}{dt}=-0.0005y\\ y\left(t\right)=y\left(0\right)\times e^{kt}\\ \left[N_2{O}_5\right]\left(t\right)=\left[N_2{O}_5\right]\left(0\right)\times e^{-0.0005t}\\ \left[N_2{O}_5\right]\left(t\right)=C\times e^{-0.0005t}\end{array}$

b.

To determine

How long will the reaction take to reduce the concentration of $N_2{O}_5$ to $90\%$ of its original value

Answer to Problem 7E

The reaction takes to reduce the concentration in $210.721\text{ seconds}$

Explanation of Solution

Given information : The question is

The chemical equation is $N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2$ takes place at ${45}^{\circ }C$ , the rate of reaction of dinitrogen pent oxide is proportional to its concentration

  $-\frac{d\left[N_2{O}_5\right]}{dt}=0.0005\left[N_2{O}_5\right]$

Calculation :

  $\begin{array}{l}\left[N_2{O}_5\right]\left(t\right)=C\times e^{-0.0005t}\\ 0.9C=C\times e^{-0.0005t}\\ 0.9=e^{-0.0005t}\\ \ln \left(0.9\right)=-0.0005t\\ \frac{\ln \left(0.9\right)}{-0.0005}=t\\ t=210.721\text{ seconds}\end{array}$