Chapter 5, Problem 5H.2E

(a)

Interpretation Introduction

Interpretation:

The value of $K$ for the given below reaction has to be calculated.

  $\frac{1}{2}NO\left(g\right)+\frac{1}{4}{O}_2\left(g\right)\rightleftharpoons \frac{1}{2}N{O}_2\left(g\right)$

Concept Introduction:

Relations between equilibrium constants:

Chemical equation	Equilibrium constant
$aA+bB\rightleftharpoons cC+dD$	$K_1$
$cC+dD\rightleftharpoons aA+bB$	$K_2=1/K_1$
$N\left(aA+bB\rightleftharpoons cC+dD\right)$	$K_3=K_1{}^N$

Answer to Problem 5H.2E

The value of $K$ for $\frac{1}{2}NO\left(g\right)+\frac{1}{4}{O}_2\left(g\right)\rightleftharpoons \frac{1}{2}N{O}_2\left(g\right)$ is $397.63.$

Explanation of Solution

The given reaction is as follows,

  $2NO\left(g\right)+O_2\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)K_1=2.5\times {10}^{10}$

By multiplying $1/4$ in the above reaction, the required reaction can be obtained.

  $\frac{1}{2}NO\left(g\right)+\frac{1}{4}{O}_2\left(g\right)\rightleftharpoons \frac{1}{2}N{O}_2\left(g\right)K_2={\left(K_1\right)}^{1/4}={\left(2.5\times {10}^{10}\right)}^{1/4}=397.63$

Therefore, the value of $K$ for $\frac{1}{2}NO\left(g\right)+\frac{1}{4}{O}_2\left(g\right)\rightleftharpoons \frac{1}{2}N{O}_2\left(g\right)$ is $397.63.$

(b)

Interpretation Introduction

Interpretation:

The value of $K$ for the given below reaction has to be calculated.

  $4N{O}_2\left(g\right)\rightleftharpoons 4NO\left(g\right)+2O_2\left(g\right)$

Concept Introduction:

Refer to part (a).

Answer to Problem 5H.2E

The value of $K$ for $4N{O}_2\left(g\right)\rightleftharpoons 4NO\left(g\right)+2O_2\left(g\right)$ is $0.024.$

Explanation of Solution

The given reaction is as follows,

  $2NO\left(g\right)+O_2\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)K_1=2.5\times {10}^{10}$

The reverse reaction of the above reaction can be written as follows,

  $2N{O}_2\left(g\right)\rightleftharpoons 2NO\left(g\right)+O_2\left(g\right)K_2=1/K_1=1/\left(2.5\times {10}^{10}\right)=4.0\times 10^{-11}$

By multiplying $2$ in the above reaction, the required reaction can be obtained.

  $\begin{array}{l}2\times \left[2N{O}_2\left(g\right)\rightleftharpoons 2NO\left(g\right)+O_2\left(g\right)\right]\\ \\ 4N{O}_2\left(g\right)\rightleftharpoons 4NO\left(g\right)+2O_2\left(g\right)K_3={\left(K_2\right)}^2={\left(\text{4}{\text{.0×10}}^{\text{-11}}\right)}^2=1.6\times 10^{-21}\end{array}$

Therefore, the value of $K$ for $4N{O}_2\left(g\right)\rightleftharpoons 4NO\left(g\right)+2O_2\left(g\right)$ is $1.6\times 10^{-21}.$

(c)

Interpretation Introduction

Interpretation:

The value of $K$ for the given below reaction has to be calculated.

  $6NO\left(g\right)+3O_2\left(g\right)\rightleftharpoons 6N{O}_2\left(g\right)$

Concept Introduction:

Refer to part (a).

Answer to Problem 5H.2E

The value of $K$ for $6NO\left(g\right)+3O_2\left(g\right)\rightleftharpoons 6N{O}_2\left(g\right)$ is $1.5625\times 10^{-31}.$

Explanation of Solution

The given reaction is as follows,

  $2NO\left(g\right)+O_2\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)K_1=2.5\times {10}^{10}$

By multiplying $3$ in the above reaction, the required reaction can be obtained.

  $\begin{array}{l}3\times \left[2NO\left(g\right)+O_2\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)\right]\\ \\ 6NO\left(g\right)+3O_2\left(g\right)\rightleftharpoons 6N{O}_2\left(g\right)K_2={\left(K_1\right)}^3={\left(2.5\times {10}^{10}\right)}^3=1.5625\times 10^{-31}\end{array}$

Therefore, the value of $K$ for $6NO\left(g\right)+3O_2\left(g\right)\rightleftharpoons 6N{O}_2\left(g\right)$ is $1.5625\times 10^{-31}.$