Chapter 21, Problem 21.124QP

(a)

Interpretation Introduction

Interpretation:

The standard enthalpy change for the given reaction involving synthesis of gas has to be calculated

Concept Introduction:

Enthalpy of reaction: ( $\Delta \text{H}°$)

The enthalpy of a reaction is calculated by subtracting the heat of formation of reactants from heat of formation of products.

$\Delta \text{H}°\text{ =}\sum \Delta \text{H}{°}_f{}_{\mathrm{Pr}\text{odcuts}}\text{ - }\sum \Delta \text{H}{°}_{f\text{ Reactants}}$

Answer to Problem 21.124QP

The standard enthalpy change for the given reaction involving synthesis of gas is $-90.2\text{ kJ}$

Explanation of Solution

Given data:

The given reaction involving synthesis of gas is:

${\text{CO(g) + 2H}}_{\text{2}}\text{(g)}\stackrel{}{\to }{\text{ CH}}_{\text{3}}\text{OH}(g)$

To Calculate: The standard enthalpy change for the given reaction involving synthesis of gas

Using the Appendix C, the heat of formation ( $\Delta \text{H}{°}_f$) of the reactants and products are written under their equation as follows,

$\begin{array}{l}{\text{ CO(g) + 2H}}_{\text{2}}\text{(g)}\stackrel{}{\to }{\text{ CH}}_{\text{3}}\text{OH}(g)\\ \Delta \text{H}{°}_f-110.50\text{ -200}\text{.7}\text{ (kJ)}\end{array}$

The $\Delta \text{H}°$ is calculated as follows,

$\begin{array}{l}\Delta \text{H}°\text{=}\sum \Delta \text{H}{°}_f{}_{\mathrm{Pr}\text{odcuts}}\text{ - }\sum \Delta \text{H}{°}_{f\text{ Reactants}}\\ =\text{[-200}\text{.7 - (-110}\text{.5)] kJ}\\ \text{= -90}\text{.2 kJ}\end{array}$

Hence, the standard enthalpy change is $-90.2\text{ kJ}$

Conclusion

The standard enthalpy change for the given reaction involving synthesis of gas is calculated as $-90.2\text{ kJ}$

(b)

Interpretation Introduction

Interpretation:

The standard enthalpy change for the given reaction involving synthesis of gas has to be calculated

Concept Introduction:

Enthalpy of reaction: ( $\Delta \text{H}°$)

The enthalpy of a reaction is calculated by subtracting the heat of formation of reactants from heat of formation of products.

$\Delta \text{H}°\text{ =}\sum \Delta \text{H}{°}_f{}_{\mathrm{Pr}\text{odcuts}}\text{ - }\sum \Delta \text{H}{°}_{f\text{ Reactants}}$

Answer to Problem 21.124QP

The standard enthalpy change for the given reaction involving synthesis of gas is $-206.2\text{ kJ}$

Explanation of Solution

Given data:

The given reaction involving synthesis of gas is:

${\text{CO(g) + 3H}}_{\text{2}}\text{(g)}\stackrel{}{\to }{\text{ CH}}_4(g){\text{ + H}}_{\text{2}}\text{O}(g)$

To Calculate: The standard enthalpy change for the given reaction involving synthesis of gas

Using the Appendix C, the heat of formation ( $\Delta \text{H}{°}_f$) of the reactants and products are written under their equation as follows,

$\begin{array}{l}{\text{ CO(g) + 3H}}_{\text{2}}\text{(g)}\stackrel{}{\to }{\text{ CH}}_4(g){\text{ + H}}_{\text{2}}\text{O}(g)\\ \Delta \text{H}{°}_f\text{ -110}\text{.5}0\text{ -74}\text{.87}\text{ -241}\text{.8}\text{ (kJ)}\end{array}$

The $\Delta \text{H}°$ is calculated as follows,

$\begin{array}{l}\Delta \text{H}°\text{=}\sum \Delta \text{H}{°}_f{}_{\mathrm{Pr}\text{odcuts}}\text{ - }\sum \Delta \text{H}{°}_{f\text{ Reactants}}\\ =\text{[(-74}\text{.87) + (-241}\text{.8) - (-110}\text{.5)] kJ}\\ \text{= -206}\text{.12 kJ}\end{array}$

Hence, the standard enthalpy change is $-206.2\text{ kJ}$

Conclusion

The standard enthalpy change for the given reaction involving synthesis of gas is calculated as $-206.2\text{ kJ}$