Chapter 14, Problem 14.83P

(a)

Interpretation Introduction

Interpretation:

The value of $\Delta U{°}_{\text{rxn}}$ for following reaction has to be determined.

  ${\text{2H}}_{\text{2}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{2H}}_{\text{2}}\text{O}\left(g\right)$

Concept Introduction:

Thermodynamics is a study of energy transfers which can be done by either heat or work. The energy transferred through work involves force. When work is positive then system gains energy while when work is negative then system loses energy. The expression to calculate work is as follows:

  $w=-P\left(V_{\text{f}}-V_{\text{i}}\right)$        (1)

Here $P$ is pressure, $V_{\text{f}}$ is the final volume, and $V_{\text{i}}$ is initial volume.

The expression of ideal gas is as follows:

  $PV=nRT$        (2)

For change in volume of gas, equation (2) can also be written as follows:

  $P\Delta V_{\text{rxn}}^{\text{o}}=\Delta n_{\text{gas}}RT$        (3)

Here, $\Delta V_{\text{rxn}}^{\text{o}}$ is change in volume of gas, $\Delta n_{\text{gas}}$ change in number of moles of gas.

Substitute $\Delta n_{\text{gas}}RT$ for $P\left(V_{\text{f}}-V_{\text{i}}\right)$ in equation (1).

  $w=-\Delta n_{\text{g}}RT$        (4)

The expression to calculate difference between numbers of moles of product and reactant $\left(\Delta n_{\text{g}}\right)$ is as follows:

  $\Delta n_{\text{g}}=n_{\text{p}}-n_{\text{r}}$        (5)

Here,

$n_{\text{r}}$ is total moles of gaseous reactant.

$n_{\text{p}}$ is total moles of gaseous product.

Answer to Problem 14.83P

The value of $\Delta U{°}_{\text{rxn}}$ for given reaction is $-569.12\text{ kJ}\cdot {\text{mol}}^{-1}$.

Explanation of Solution

The given chemical equation is as follows:

  ${\text{2H}}_{\text{2}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{2H}}_{\text{2}}\text{O}\left(g\right)$        (6)

Total moles gaseous reactant $\left(n_{\text{r}}\right)$ are3 moles.

Total moles of gaseous product $\left(n_{\text{p}}\right)$ are2 moles.

Substitute 2 for $n_{\text{p}}$ and 3 for $n_{\text{r}}$ in equation (5).

  $\begin{array}{c}\Delta n_{\text{g}}=2-3\\ =-1\end{array}$

Substitute 1 for $\Delta n_{\text{g}}$ in equation (4).

  $\begin{array}{c}w=-\left(-1\right)RT\\ =+RT\end{array}$

The expression to calculate value of $\Delta U{°}_{\text{rxn}}$ is as follows:

  $\Delta U{°}_{\text{rxn}}=\Delta H{°}_{\text{rxn}}+w$        (7)

Substitute $+RT$ for $w$ in equation (7).

  $\Delta U{°}_{\text{rxn}}=\Delta H{°}_{\text{rxn}}+RT$        (8)

Substitute $-571.6\text{ kJ}\cdot {\text{mol}}^{-1}$ for $\Delta H{°}_{\text{rxn}}$, $8.314\text{ J}\cdot {\text{mol}}^{-1}\cdot {\text{K}}^{-1}$ for $R$, and $298\text{ K}$ for $T$ in equation (8).

  $\begin{array}{c}\Delta U{°}_{\text{rxn}}=\left(-571.6\text{ kJ}\cdot {\text{mol}}^{-1}\right)+\left(8.314\text{ J}\cdot {\text{mol}}^{-1}\cdot {\text{K}}^{-1}\right)\left(298\text{ K}\right)\left(\frac{1\text{ kJ}}{{10}^3\text{ J}}\right)\\ =-569.12\text{ kJ}\cdot {\text{mol}}^{-1}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The value of $\Delta U{°}_{\text{rxn}}$ for following reaction has to be determined.

  $\text{Sn}\left(s\right)+2{\text{Cl}}_2\left(g\right)\to {\text{SnCl}}_4\left(l\right)$

Concept Introduction:

Refer to part (a).

Answer to Problem 14.83P

The value of $\Delta U{°}_{\text{rxn}}$ for given reaction is $-506.34\text{ kJ}\cdot {\text{mol}}^{-1}$.

Explanation of Solution

The given chemical equation is as follows:

  $\text{Sn}\left(s\right)+2{\text{Cl}}_2\left(g\right)\to {\text{SnCl}}_4\left(l\right)$        (9)

Total moles gaseous reactant $\left(n_{\text{r}}\right)$ are 3 moles.

Total moles of gaseous product $\left(n_{\text{p}}\right)$ are1 moles.

Substitute 1 for $n_{\text{p}}$ and 3 for $n_{\text{r}}$ in equation (5).

  $\begin{array}{c}\Delta n_{\text{g}}=1-3\\ =-2\end{array}$

Substitute $-2$ for $\Delta n_{\text{g}}$ in equation (4).

  $\begin{array}{c}w=-\left(-2\right)RT\\ =+2RT\end{array}$

The expression to calculate value of $\Delta U{°}_{\text{rxn}}$ is as follows:

  $\Delta U{°}_{\text{rxn}}=\Delta H{°}_{\text{rxn}}+w$        (7)

Substitute $+2RT$ for $w$ in equation (7).

  $\Delta U{°}_{\text{rxn}}=\Delta H{°}_{\text{rxn}}+2RT$        (10)

Substitute $-511.3\text{ kJ}\cdot {\text{mol}}^{-1}$ for $\Delta H{°}_{\text{rxn}}$, $8.314\text{ J}\cdot {\text{mol}}^{-1}\cdot {\text{K}}^{-1}$ for $R$, and $298\text{ K}$ for $T$ in equation (10).

  $\begin{array}{c}\Delta U{°}_{\text{rxn}}=\left(-511.3\text{ kJ}\cdot {\text{mol}}^{-1}\right)+2\left(8.314\text{ J}\cdot {\text{mol}}^{-1}\cdot {\text{K}}^{-1}\right)\left(298\text{ K}\right)\left(\frac{1\text{ kJ}}{{10}^3\text{ J}}\right)\\ =-506.34\text{ kJ}\cdot {\text{mol}}^{-1}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The value of $\Delta U{°}_{\text{rxn}}$ for following reaction has to be determined.

  ${\text{H}}_2\left(g\right)+{\text{Cl}}_2\left(g\right)\to 2\text{HCl}\left(g\right)$

Concept Introduction:

Refer to part (a).

Answer to Problem 14.83P

The value of $\Delta U{°}_{\text{rxn}}$ for given reaction is $-184.6\text{ kJ}\cdot {\text{mol}}^{-1}$.

Explanation of Solution

The given chemical equation is as follows:

  ${\text{H}}_2\left(g\right)+{\text{Cl}}_2\left(g\right)\to 2\text{HCl}\left(g\right)$        (11)

Total moles gaseous reactant $\left(n_{\text{r}}\right)$ are 2 moles.

Total moles of gaseous product $\left(n_{\text{p}}\right)$ are 2 moles.

Substitute 2 for $n_{\text{p}}$ and 2 for $n_{\text{r}}$ in equation (5).

  $\begin{array}{c}\Delta n_{\text{g}}=2-2\\ =0\end{array}$

Substitute 0 for $\Delta n_{\text{g}}$ in equation (4).

  $\begin{array}{c}w=-\left(0\right)RT\\ =0\end{array}$

The expression to calculate value of $\Delta U{°}_{\text{rxn}}$ is as follows:

  $\Delta U{°}_{\text{rxn}}=\Delta H{°}_{\text{rxn}}+w$        (7)

Substitute 0 for $w$ in equation (7).

  $\begin{array}{c}\Delta U{°}_{\text{rxn}}=\Delta H{°}_{\text{rxn}}+0\\ =\Delta H{°}_{\text{rxn}}\end{array}$

Hence value of $\Delta U{°}_{\text{rxn}}$ is equal to value of $\Delta H{°}_{\text{rxn}}$ and its value is $-184.6\text{ kJ}\cdot {\text{mol}}^{-1}$.