Chapter 21, Problem 53QAP

Consider the reaction
$4{\text{NH}}_3\left(g\right)+5{\text{O}}_2\left(g\right)\to 4\text{NO}\left(g\right)+6{\text{H}}_2\text{O}\left(g\right)$
(a) Calculate ?H° for this reaction. Is it exothermic or endothermic?
(b) Would you expect ?S° to be positive or negative? Calculate ?S°.
(c) Is the reaction spontaneous at 25°C and 1 atm?
(d) At what temperature, if any, is the reaction at equilibrium at 1 atm pressure?

Interpretation Introduction

(a)

Interpretation:

To calculate $\Delta H^0$ for the below reaction and to determine reaction is endothermic or exothermic.

${\text{4NH}}_{\text{3}}\left(\text{g}\right)+{\text{ 5O}}_{\text{2}}\left(\text{g}\right)\to \text{4NO}\left(\text{g}\right)+{\text{ 6H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Concept introduction:

A reaction is endothermic if it gains energy from surroundings that is, $\Delta H^0>0$ and a reaction is exothermic if it releases energy to surroundings that is $\Delta H^0<0$.

Answer to Problem 53QAP

$\Delta H^0<0$ and reaction is exothermic.

Explanation of Solution

From appendix (1),

$\begin{array}{l}\Delta H^0{}_f\left(\text{NO}\right)=90.2\text{kJ}\\ \Delta H^0{}_f\left({\text{H}}_{\text{2}}\text{O}\right)=-241.8\text{kJ}\\ \Delta H^0{}_f\left({\text{O}}_2\right)=0\text{kJ}\\ \Delta H^0{}_f\left({\text{NH}}_{\text{3}}\right)=-46.1\text{kJ}\end{array}$

$\begin{array}{c}\Delta H^0=\Delta H^0{}_f\left(\text{Product}\right)-\Delta H^0{}_f\left(\text{Reactant}\right)\\ =\left(4\times \left(90.2\text{kJ/mol}\right)+6\times \left(-241.8\right)\text{kJ/mol}\right)-\left(4\times \left(-46.1\text{kJ/mol}\right)+5\times 0\text{kJ/mol}\right)\\ =-1090\text{kJ/mol+184}\text{.4 kJ/mol}\\ =-905.6\text{kJ/mol}\end{array}$

Hence, value of $\Delta H^0=-905.6\text{kJ/mol}$

Since, $\Delta H^0<0$, above reaction is exothermic.

Interpretation Introduction

(b)

Interpretation:

To calculate $\Delta S^0$ for the below reaction and to determine $\Delta S^0$ is positive or negative.

${\text{4NH}}_{\text{3}}\left(\text{g}\right)+{\text{ 5O}}_{\text{2}}\left(\text{g}\right)\to \text{4NO}\left(\text{g}\right)+{\text{ 6H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Concept introduction:

Standard entropy is calculated from difference of entropy of product and reactant.

$\Delta S^0=\Delta S^0\left(\text{Product}\right)-\Delta S^0\left(\text{Reactant}\right)$

Where, $\Delta S^0$ is the standard entropy.

Answer to Problem 53QAP

$\Delta S^0>0$, value of $\Delta S^0$ is positive.

Explanation of Solution

From appendix (1),

$\begin{array}{l}{S}^0\left(\text{NO}\right)=0.2107\text{kJ/K}\cdot \text{mol}\\ S^0\left({\text{H}}_{\text{2}}\text{O}\right)=0.1887\text{kJ/K}\cdot \text{mol}\\ S^0\left({\text{O}}_2\right)=0.2050\text{kJ/K}\cdot \text{mol}\\ S^0\left({\text{NH}}_{\text{3}}\right)=0.1923\text{kJ/K}\cdot \text{mol}\end{array}$

$\begin{array}{c}\Delta S^0=\Delta S^0\left(\text{Product}\right)-\Delta S^0\left(\text{Reactant}\right)\\ =\left(4\times 0.2107\text{kJ/K}\cdot \text{mol}+6\times 0.1887\text{kJ/K}\cdot \text{mol}\right)-\left(4\times 0.1923\text{kJ/K}\cdot \text{mol}+5\times 0.2050\text{kJ/K}\cdot \text{mol}\right)\\ =\left(0.8428\text{kJ/K}\cdot \text{mol}+1.1322\text{kJ/K}\cdot \text{mol}\right)-\left(0.7692\text{kJ/K}\cdot \text{mol}+1.025\text{kJ/K}\cdot \text{mol}\right)\\ =1.975\text{kJ/K}\cdot \text{mol}-1.7942\text{kJ/K}\cdot \text{mol}\end{array}$$\Delta S^0=0.1808\text{kJ/K}$

Hence, $\Delta S^0=0.1808\text{kJ/K}$.

Interpretation Introduction

(c)

Interpretation:

To find the below reaction is spontaneous or not at $25°\text{C and 1 atm}$.

${\text{4NH}}_{\text{3}}\left(\text{g}\right)+{\text{ 5O}}_{\text{2}}\left(\text{g}\right)\to \text{4NO}\left(\text{g}\right)+{\text{ 6H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Concept introduction:

A reaction is called spontaneous if the value of Standard Gibbs Free energy is less than zero.

Answer to Problem 53QAP

Reaction is spontaneous.

Explanation of Solution

From subpart (a), $\Delta H^0=-905.6\text{kJ/mol}$

From subpart (b), $\Delta S^0=0.1808\text{kJ/K}$

$\begin{array}{c}T=25°\text{C}\\ \text{=}25+273\text{K}\\ \text{=298}\text{K}\end{array}$

From Gibbs Helmholtz equation,

$\begin{array}{c}\Delta G^0=\Delta H^0-T\Delta S^0\\ =-905.6\text{kJ}-\left(298\text{K}\times 0.1808\text{kJ/K}\right)\\ =-905.6\text{kJ}-53.8784\text{kJ}\\ \text{=}-\text{959}\text{.48}\text{kJ}\end{array}$

Since, $\Delta G^0<0$, nature of the reaction is spontaneous.

Interpretation Introduction

(d)

Interpretation:

To find the temperature at which following reaction is in equilibrium.

${\text{4NH}}_{\text{3}}\left(\text{g}\right)+{\text{ 5O}}_{\text{2}}\left(\text{g}\right)\to \text{4NO}\left(\text{g}\right)+{\text{ 6H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Concept introduction:

Gibbs Helmholtz equation states relation between Gibbs free energy, enthalpy, temperature and entropy.

$\Delta G^0=\Delta H^0-T\Delta S^0$

Where,

$\Delta G^0$ is standard Gibbs Free energy, $\Delta H^0$ is standard enthalpy and $\Delta S^0$ is standard entropy.

Answer to Problem 53QAP

T = $-5009\text{K}$, this value of temperature is not possible.

Hence, the reaction can’t be in equilibrium.

Explanation of Solution

For a reaction to attain equilibrium, $\Delta G^0=0$

In that case, in Gibbs Helmholtz equation,

$\Delta G^0=\Delta H^0-T\Delta S^0$

As, $\Delta G^0=0$

$\Delta H^0=T\Delta S^0$

$T=\frac{\Delta H^0}{\Delta S^0}$ …… (1)

$\Delta H^0=-905.6\text{kJ/mol}$

$\Delta S^0=0.1808\text{kJ/K}$

Above values are to be put in equation (1).

$\begin{array}{c}T=\frac{\Delta H^0}{\Delta S^0}\\ =\frac{-905.6\text{kJ}}{0.1808\text{kJ/K}}\\ =-5009\text{K}\end{array}$

This value of temperature is not possible.

Hence, the reaction can’t be in equilibrium.