Chapter 14, Problem 1DE

(a)

Interpretation Introduction

To determine: The standard enthalpy change of the given reaction.

Answer to Problem 1DE

Solution: The standard enthalpy change of the given reaction is $-217.52\text{kJ}$ .

Explanation of Solution

Given The balanced chemical equation for the given reaction is,

$\text{CaO}\left(s\right)+2\text{HCl}\left(g\right)\to {\text{CaCl}}_2\left(s\right)+{\text{H}}_2\text{O}\left(g\right)$

The standard enthalpy change of the given reaction is calculated by the formula,

$\begin{array}{c}\Delta H^{\circ }={\displaystyle \sum n\Delta H^{\circ }\left(\text{products}\right)}-{\displaystyle \sum m\Delta H^{\circ }\left(\text{reactants}\right)}\\ =\left(\Delta H_{{\text{CaCl}}_2\left(s\right)}^{\circ }+\Delta H_{{\text{H}}_2\text{O}\left(g\right)}^{\circ }\right)-\left(2\Delta H_{\text{HCl}\left(g\right)}^{\circ }+\Delta H_{\text{CaO}\left(s\right)}^{\circ }\right)\end{array}$

Substitute the values of $\Delta H_{{\text{CaCl}}_2\left(s\right)}^{\circ }$ , $\Delta H_{\text{HCl}\left(g\right)}^{\circ }$ , $\Delta H_{{\text{H}}_2\text{O}\left(g\right)}^{\circ }$ and $\Delta H_{\text{CaO}\left(s\right)}^{\circ }$ from appendix C in the above equation.

$\begin{array}{c}\Delta H^{\circ }=\left(1\text{mol}\times -795.8\text{kJ/mol}+1\text{mol}\times -241.82\text{kJ/mol}\right)\\ -\left(2\text{mol}\times -92.30\text{kJ/mol}+1\text{mol}\times -635.5\text{kJ/mol}\right)\\ =-217.52\text{kJ}\end{array}$

Conclusion

The standard enthalpy change of the given reaction is $-217.52\text{kJ}$ .

(b)

Interpretation Introduction

To determine: The standard enthalpy change of the given reaction.

Answer to Problem 1DE

Solution: The standard enthalpy change of the given reaction is $-556.72\text{kJ}$ .

Explanation of Solution

Given The balanced chemical equation of the given reaction is,

$\text{4FeO}\left(s\right)+{\text{O}}_2\left(g\right)\to 2{\text{Fe}}_2{\text{O}}_3\left(s\right)$

The standard enthalpy change of the given reaction is calculated by the formula,

$\begin{array}{c}\Delta H^{\circ }={\displaystyle \sum n\Delta H^{\circ }\left(\text{products}\right)}-{\displaystyle \sum m\Delta H^{\circ }\left(\text{reactants}\right)}\\ =2\Delta H_{{\text{Fe}}_2{\text{O}}_3\left(s\right)}^{\circ }-\left(4\Delta H_{\text{FeO}\left(s\right)}^{\circ }+\Delta H_{{\text{O}}_2\left(g\right)}^{\circ }\right)\end{array}$

Substitute the values of $\Delta H_{{\text{Fe}}_2{\text{O}}_3\left(s\right)}^{\circ }$ , $\Delta H_{\text{FeO}\left(s\right)}^{\circ }$ and $\Delta H_{{\text{O}}_2\left(g\right)}^{\circ }$ from appendix C in the above equation.

$\begin{array}{c}\Delta H^{\circ }=2\text{mol}\times -822.16\text{kJ/mol}-\left(4\text{mol}\times -271.9\text{kJ/mol}+1\text{mol}\times 0.0\text{kJ/mol}\right)\\ =-556.72\text{kJ}\end{array}$

Conclusion

The standard enthalpy change of the given reaction is $-556.72\text{kJ}$ .

(c)

Interpretation Introduction

To determine: The standard enthalpy change of the given reaction.

Answer to Problem 1DE

Solution: The standard enthalpy change of the given reaction is $84.97\text{kJ}$ .

Explanation of Solution

Given The balanced chemical equation for the given reaction is,

$\text{2CuO}\left(s\right)+\text{NO}\left(g\right)\to {\text{Cu}}_2\text{O}\left(s\right)+{\text{NO}}_2\left(g\right)$

The standard enthalpy change of the given reaction is calculated by the formula,

$\begin{array}{c}\Delta H^{\circ }={\displaystyle \sum n\Delta H^{\circ }\left(\text{products}\right)}-{\displaystyle \sum m\Delta H^{\circ }\left(\text{reactants}\right)}\\ =\left(\Delta H_{{\text{Cu}}_2\text{O}\left(s\right)}^{\circ }+\Delta H_{{\text{NO}}_2\left(g\right)}^{\circ }\right)-\left(2\Delta H_{\text{CuO}\left(s\right)}^{\circ }+\Delta H_{\text{NO}\left(g\right)}^{\circ }\right)\end{array}$

Substitute the values of $\Delta H_{{\text{Cu}}_2\text{O}\left(s\right)}^{\circ }$ , $\Delta H_{{\text{NO}}_2\left(g\right)}^{\circ }$ , $\Delta H_{\text{CuO}\left(s\right)}^{\circ }$ and $\Delta H_{\text{NO}\left(g\right)}^{\circ }$ from appendix C in the above equation.

$\begin{array}{c}\Delta H^{\circ }=\left(1\text{mol}\times -170.7\text{kJ/mol}+1\text{mol}\times 33.84\text{kJ/mol}\right)\\ -\left(2\text{mol}\times -156.1\text{kJ/mol}+1\text{mol}\times 90.37\text{kJ/mol}\right)\\ =84.97\text{kJ}\end{array}$

Conclusion

The standard enthalpy change of the given reaction is $84.97\text{kJ}$ .

(d)

Interpretation Introduction

To determine: The standard enthalpy change of the given reaction.

Answer to Problem 1DE

Solution: The standard enthalpy change of the given reaction is $-196.10\text{kJ}$ .

Explanation of Solution

Given The balanced chemical equation for the given reaction is,

${\text{4NH}}_3\left(g\right)+{\text{O}}_2\left(g\right)\to 2{\text{N}}_2{\text{H}}_4\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$

The standard enthalpy change of the given reaction is calculated by the formula,

$\begin{array}{c}\Delta H^{\circ }={\displaystyle \sum n\Delta H^{\circ }\left(\text{products}\right)}-{\displaystyle \sum m\Delta H^{\circ }\left(\text{reactants}\right)}\\ =\left(2\Delta H_{{\text{N}}_2{\text{H}}_4\left(g\right)}^{\circ }+2\Delta H_{{\text{H}}_2\text{O}\left(g\right)}^{\circ }\right)-\left(4\Delta H_{{\text{NH}}_3\left(g\right)}^{\circ }+\Delta H_{{\text{O}}_2\left(g\right)}^{\circ }\right)\end{array}$

Substitute the values of $\Delta H_{{\text{N}}_2{\text{H}}_4\left(g\right)}^{\circ }$ , $\Delta H_{{\text{H}}_2\text{O}\left(g\right)}^{\circ }$ , $\Delta H_{{\text{H}}_2\text{O}\left(g\right)}^{\circ }$ and $\Delta H_{{\text{NH}}_3\left(g\right)}^{\circ }$ from appendix C in the above equation.

$\begin{array}{c}\Delta H^{\circ }=\left(2\text{mol}\times 95.40\text{kJ/mol}+2\text{mol}\times -285.83\text{kJ/mol}\right)\\ -\left(4\text{mol}\times -46.19\text{kJ/mol}+1\text{mol}\times 0.0\text{kJ/mol}\right)\\ =-196.10\text{kJ}\end{array}$

Conclusion

The standard enthalpy change of the given reaction is $-196.10\text{kJ}$ .