Chapter 2, Problem 2C.5BE

(i)

Interpretation Introduction

Interpretation: The values of ${\Delta }_{\text{r}}H°$ and ${\Delta }_{\text{r}}U°$ has to be calculated for the given reaction.

Concept introduction: Enthalpy is the internal heat content of the system.  Under constant pressure, a system can do expansion work under the effect of energy supplied as heat known as enthalpy change of the system.  Therefore, the measure of amount of heat supplied to the system gives the enthalpy change of the system.  The total amount of energy possessed within the given system is known as internal energy of the system. It is denoted by $U$.

Answer to Problem 2C.5BE

The value of ${\Delta }_{\text{r}}H°$ for the reaction (3) is $\underset{\_}{-589.56kJmo{l}^{-1}}$.  The value of ${\Delta }_{\text{r}}U°$ for the reaction (3) is $\underset{\_}{-587.08kJmo{l}^{-1}}$.

Explanation of Solution

The given reactions are:

    ${\text{H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{I}}_{\text{2}}\left(\text{g}\right)\to \text{2HI}\left(\text{g}\right)$                                                                           (1)

    ${\text{2H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{2H}}_2\text{O}\left(\text{g}\right)$                                                                      (2)

    $\text{4HI}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{2I}}_2\left(\text{g}\right)\text{+}{\text{2H}}_{\text{2}}\text{O}\left(\text{g}\right)$                                                        (3)

The given value of ${\Delta }_{\text{r}}H°$ for the reaction (1) is $52.96\text{kJ}{\text{mol}}^{-1}$ and the given value of ${\Delta }_{\text{r}}H°$ for the reaction (2) is $-483.64\text{kJ}{\text{mol}}^{-1}$.

To calculate the ${\Delta }_{\text{r}}H°$ value for the reaction (3), multiply reaction (1) by $2$ and reverse it as shown below.

    $2\times \left({\text{H}}_{\text{2}}\text{(g)}\text{+}{\text{I}}_{\text{2}}\text{(g)}\to \text{2HI(g)}\right)$

    $\text{4HI(g)}\to 2{\text{H}}_{\text{2}}\text{(g)}\text{+}{\text{2I}}_{\text{2}}\text{(g)}$                                                                            (4)

The value of ${\Delta }_{\text{r}}H°$ for the reaction (4) becomes $2\times \left(-52.96\text{kJ}{\text{mol}}^{-1}\right)=-105.92\text{kJ}{\text{mol}}^{-1}$

Add reaction (4) to reaction (2).

    $\begin{array}{l}\text{4HI(g)}\to 2{\text{H}}_{\text{2}}\text{(g)}\text{+}{\text{2I}}_{\text{2}}\text{(g)}\\ {\text{2H}}_{\text{2}}\text{(g)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\to {\text{2H}}_2\text{O(g)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ \text{4HI(g)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\to {\text{2I}}_2\text{(g)}\text{+}{\text{2H}}_{\text{2}}\text{O}\text{(g)}\end{array}$

To calculate the ${\Delta }_{\text{r}}H°$ value for the reaction (3), use the enthalpy of reaction (4) and (2).

    $\begin{array}{c}{\Delta }_{\text{r}}H°=-105.92\text{kJ}{\text{mol}}^{-1}+\left(-483.64\right)\text{kJ}{\text{mol}}^{-1}\\ =\underset{\_}{-589.56kJmo{l}^{-1}}\end{array}$

Hence, the value of ${\Delta }_{\text{r}}H°$ for the reaction (3) is $\underset{\_}{-589.56kJmo{l}^{-1}}$.

The change in number of moles of the reaction (3) is calculated by the formula,

  $\text{Number}\text{of}\text{moles,}\Delta n_{\text{g}}=\text{Moles}\text{of}\text{gaseous}\text{product}-\text{Moles}\text{of}\text{gaseous}\text{reactant}$

Hence, the number of moles for the reaction (3) is,

  $\begin{array}{c}\Delta n_{\text{g}}=\left(2+2\right)-\left(4+1\right)\\ =-1\end{array}$

The expression to calculate the value of ${\Delta }_{\text{r}}U°$ for the reaction (3) is,

    ${\Delta }_{\text{r}}U°={\Delta }_{\text{r}}H°-\Delta n_{\text{g}}RT$

Substitute the values of ${\Delta }_{\text{r}}H°$, $R$, $\Delta n_{\text{g}}$ and $T=298\text{K}$ in the above equation to calculate the value of ${\Delta }_{\text{r}}U°$.

  $\begin{array}{c}{\Delta }_{\text{r}}U°=-589.56\text{kJ}{\text{mol}}^{-1}-\left(-1\times 8.314\times {10}^{-3}\text{kJ}{\text{K}}^{-1}{\text{mol}}^{-1}\times 298\text{K}\right)\\ =-589.56\text{kJ}{\text{mol}}^{-1}-\left(-2477.572\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}\right)\\ =\underset{\_}{-587.08kJmo{l}^{-1}}\end{array}$

Hence, the value of ${\Delta }_{\text{r}}U°$ for the reaction (3) is $\underset{\_}{-587.08kJmo{l}^{-1}}$.

(ii)

Interpretation Introduction

Interpretation: The value of ${\Delta }_{\text{f}}H°$ for both $\text{HI}\left(\text{g}\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ has to be calculated.

Concept introduction: Enthalpy is the internal heat content of the system.  Under constant pressure, a system can do expansion work under the effect of energy supplied as heat known as enthalpy change of the system.  Therefore, the measure of amount of heat supplied to the system gives the enthalpy change of the system.  The total amount of energy possessed within the given system is known as internal energy of the system. It is denoted by $U$.

Answer to Problem 2C.5BE

The value of ${\Delta }_{\text{f}}H°$ for $\text{HI}\left(\text{g}\right)$ is ${\underset{\_}{26.48kJmol}}^{-1}$.  The value of ${\Delta }_{\text{f}}H°$ for ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $-{\underset{\_}{241.82kJmol}}^{-1}$.

Explanation of Solution

The given reactions are:

    ${\text{H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{I}}_{\text{2}}\left(\text{g}\right)\to \text{2HI}\left(\text{g}\right)$                                                                           (1)

    ${\text{2H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{2H}}_2\text{O}\left(\text{g}\right)$                                                                      (2)

    $\text{4HI}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{2I}}_2\left(\text{g}\right)\text{+}{\text{2H}}_{\text{2}}\text{O}\left(\text{g}\right)$                                                        (3)

The given value of ${\Delta }_{\text{r}}H°$ for the reaction (1) is $52.96\text{kJ}{\text{mol}}^{-1}$ and the given value of ${\Delta }_{\text{r}}H°$ for the reaction (2) is $-483.64\text{kJ}{\text{mol}}^{-1}$.

The value of ${\Delta }_{\text{f}}H°$ for $\text{HCl}\text{(g)}$ is calculate by the expression,

  ${\Delta }_{\text{f}}H°=\frac{{\Delta }_{\text{r}}H°}{2}$

The value of ${\Delta }_{\text{r}}H°$ for the reaction (1) is used to calculate the value of ${\Delta }_{\text{f}}H°$ for $\text{HI}\left(\text{g}\right)$.  The value of ${\Delta }_{\text{r}}H°$ for the reaction (1) is $52.96\text{kJ}{\text{mol}}^{-1}$.

From the above expression, the value of ${\Delta }_{\text{f}}H°$ for $\text{HI}\left(\text{g}\right)$ is,

    $\begin{array}{c}{\Delta }_{\text{f}}H°=\frac{52.96\text{kJ}{\text{mol}}^{-1}}{2}\\ ={\underset{\_}{26.48kJmol}}^{-1}\end{array}$

Hence, the value of ${\Delta }_{\text{f}}H°$ for $\text{HI}\left(\text{g}\right)$ is ${\underset{\_}{26.48kJmol}}^{-1}$.

The value of ${\Delta }_{\text{f}}H°$ for ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is calculate by the expression,

  ${\Delta }_{\text{f}}H°=\frac{{\Delta }_{\text{r}}H°}{2}$

The value of ${\Delta }_{\text{r}}H°$ for the reaction (1) is used to calculate the value of ${\Delta }_{\text{f}}H°$ for ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$.  The value of ${\Delta }_{\text{r}}H°$ for the reaction (2) is $-483.64\text{kJ}{\text{mol}}^{-1}$.

From the above expression, the value of ${\Delta }_{\text{f}}H°$ for ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is,

    $\begin{array}{c}{\Delta }_{\text{f}}H°=\frac{-483.64\text{kJ}{\text{mol}}^{-1}}{2}\\ =-{\underset{\_}{241.82kJmol}}^{-1}\end{array}$

Hence, the value of ${\Delta }_{\text{f}}H°$ for ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $-{\underset{\_}{241.82kJmol}}^{-1}$.