Chapter 6, Problem 6.100P

(a)

Interpretation Introduction

Interpretation:

The standard enthalpies of reaction (1) and (2) are to be determined.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\sum m\Delta H_{\text{f (products)}}^{°}-\sum m\Delta H_{\text{f (reactants)}}^{°}$

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

Answer to Problem 6.100P

The standard enthalpies of reaction (1) and (2) are $–657.0\text{kJ}$ and $32.9\text{kJ}$ respectively.

Explanation of Solution

The balanced chemical equation for the reaction(3) is as follows:

${\text{SiCl}}_4\left(g\right)+2{\text{H}}_2\text{O}\left(g\right)\to {\text{SiO}}_2\left(s\right)+4\text{HCl}\left(g\right)$

The formula to calculate the standard enthalpy of reaction (3) $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\Delta H_{\text{f}}^{°}\left[{\text{SiO}}_2\left(s\right)\right]+4\Delta H_{\text{f}}^{°}\left[\text{HCl}\left(g\right)\right]\right\}\\ -\left\{\Delta H_{\text{f}}^{°}\left[{\text{SiCl}}_4\left(g\right)\right]+2\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]\right\}\end{array}\right]$ (1)

Rearrange equation (1) to calculate $\Delta H_{\text{f}}^{°}$ of ${\text{SiCl}}_4\left(g\right)$ is as follows:

$\Delta H_{\text{f}}^{°}\left[{\text{SiCl}}_4\left(g\right)\right]=\left[\begin{array}{l}\left\{\Delta H_{\text{f}}^{°}\left[{\text{SiO}}_2\left(s\right)\right]+4\Delta H_{\text{f}}^{°}\left[\text{HCl}\left(g\right)\right]\right\}-\\ \left(\Delta H_{\text{rxn}}^{°}+2\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]\right)\end{array}\right]$ (2)

Substitute $-910.9\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{SiO}}_2\left(s\right)\right]$, $-92.31\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[\text{HCl}\left(g\right)\right]$ and $-139.5\text{kJ}$ for $\Delta H_{\text{rxn}}^{°}$ and $-241.826\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]$ in the equation (2).

$\begin{array}{c}\left(1\text{mol}\right)\Delta H_{\text{f}}^{°}\left[{\text{SiCl}}_4\left(g\right)\right]=\left[\begin{array}{l}\left(\text{1 mol}\right)\left(-910.9\text{kJ/mol}\right)+\left(\text{4 mol}\right)\left(-92.31\text{kJ/mol}\right)-\\ \left(\left(-139.5\text{kJ}\right)+\left(\text{2 mol}\right)\left(-241.826\text{kJ/mol}\right)\right)\end{array}\right]\\ \Delta H_{\text{f}}^{°}\left[{\text{SiCl}}_4\left(g\right)\right]=\frac{-656.988\text{kJ}}{\text{1}\text{mol}}\\ =-656.988\text{kJ/mol}\end{array}$

The balanced chemical equation for the reaction(1) is as follows:

$\text{Si}\left(s\right)+2{\text{Cl}}_2\left(g\right)\to {\text{SiCl}}_4\left(g\right)$

The formula to calculate the standard enthalpy for first reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\Delta H_{\text{f}}^{°}\left[{\text{SiCl}}_4\left(g\right)\right]\right\}\\ -\left\{1\Delta H_{\text{f}}^{°}\left[\text{Si}\left(s\right)\right]+2\Delta H_{\text{f}}^{°}\left[{\text{Cl}}_2\left(g\right)\right]\right\}\end{array}\right]$ (3)

Substitute $-656.988\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{SiCl}}_4\left(g\right)\right]$, 0 for $\Delta H_{\text{f}}^{°}\left[\text{Si}\left(s\right)\right]$ and 0 for $\Delta H_{\text{f}}^{°}\left[{\text{Cl}}_2\left(g\right)\right]$ in the equation (3).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\left(1\text{mol}\right)\left(-656.988\text{kJ/mol}\right)\right\}\\ -\left\{\left(1\text{mol}\right)\left(0\right)+\left(2\text{mol}\right)\left(0\right)\right\}\end{array}\right]\\ =-656.988\text{kJ}\\ \approx –657.0\text{kJ}\end{array}$

The balanced chemical equation for the reaction(2) is as follows:

${\text{SiO}}_2\left(s\right)+2\text{C}\left(\text{graphite}\right)+2{\text{Cl}}_2\left(g\right)\to {\text{SiCl}}_4\left(g\right)+2\text{CO}\left(g\right)$

The formula to calculate the standard enthalpy for second reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\Delta H_{\text{f}}^{°}\left[{\text{SiCl}}_4\left(g\right)\right]+2\Delta H_{\text{f}}^{°}\left[\text{CO}\left(g\right)\right]\right\}\\ -\left\{1\Delta H_{\text{f}}^{°}\left[{\text{SiO}}_2\left(s\right)\right]+2\Delta H_{\text{f}}^{°}\left[\text{C}\left(\text{graphite}\right)\right]+2\Delta H_{\text{f}}^{°}\left[{\text{Cl}}_2\left(g\right)\right]\right\}\end{array}\right]$ (4)

Substitute $-656.988\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{SiCl}}_4\left(g\right)\right]$, $-110.5\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[\text{CO}\left(g\right)\right]$, $-910.9\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{SiO}}_2\left(s\right)\right]$, 0 for $\Delta H_{\text{f}}^{°}\left[\text{C}\left(\text{graphite}\right)\right]$ and 0 for $\Delta H_{\text{f}}^{°}\left[{\text{Cl}}_2\left(g\right)\right]$ in the equation (4).

$\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\left(1\text{mol}\right)\left(-656.988\text{kJ/mol}\right)+\left(2\text{mol}\right)\left(-110.5\text{kJ/mol}\right)\right\}\\ -\left\{\left(1\text{mol}\right)\left(-910.9\text{kJ/mol}\right)+\left(2\text{mol}\right)\left(0\right)+\left(2\text{mol}\right)\left(0\right)\right\}\end{array}\right]\\ =32.912\text{kJ}\\ \approx 32.9\text{kJ}\end{array}$

Conclusion

The standard enthalpies of reaction (1) and (2) are $–657.0\text{kJ}$ and $32.9\text{kJ}$ respectively.

(b)

Interpretation Introduction

Interpretation:

$\Delta H_{\text{rxn}}^{\text{°}}$ for the fourth reaction that is sum of the reaction(2) and (3) is to be determined.

Concept introduction:

Hess’s law is used to calculate the enthalpy change of an overall reaction that can be derived as a sum of two or more reaction. According to Hess’s law $\Delta H$ of an overall reaction is equal to the sum of the enthalpy change for each individual reaction. $\Delta H_{\text{overall rxn}}=\Delta H_1+\Delta H_2+\mathrm{.......}+\Delta H_{\text{n}}$

Enthalpy is a state function so the value depends upon the initial state and final state not on the path so $\Delta H$ of an overall reaction can be calculated by the addition or subtraction of the individual steps whose $\Delta H$ is known.

Answer to Problem 6.100P

$\Delta H_{\text{rxn}}^{\text{°}}$ for the fourth reaction is $-106.6\text{kJ}$.

Explanation of Solution

The enthalpy change of the following reaction is $\Delta H_2^{°}$

${\text{SiO}}_2\left(s\right)+2\text{C}\left(\text{graphite}\right)+2{\text{Cl}}_2\left(g\right)\to {\text{SiCl}}_4\left(g\right)+2\text{CO}\left(g\right)$ (5)

The enthalpy change of the following reaction is $\Delta H_3^{°}$.

${\text{SiCl}}_4\left(g\right)+2{\text{H}}_2\text{O}\left(g\right)\to {\text{SiO}}_2\left(s\right)+4\text{HCl}\left(g\right)$ (6)

Add equation (5) and (6).

$\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{SiO}}_2\left(s\right)+2\text{C}\left(\text{graphite}\right)+2{\text{Cl}}_2\left(g\right)\to {\text{SiCl}}_4\left(g\right)+2\text{CO}\left(g\right)\\ {\text{SiCl}}_4\left(g\right)+2{\text{H}}_2\text{O}\left(g\right)\to {\text{SiO}}_2\left(s\right)+4\text{HCl}\left(g\right)\end{array}}\\ 2\text{C}\left(\text{graphite}\right)+2{\text{Cl}}_2\left(g\right)+{\text{2H}}_2\text{O}\left(g\right)\to 2\text{CO}\left(g\right)+4\text{HCl}\left(g\right)\end{array}$ (8)

The enthalpy change of the final reaction (8) is $\Delta H_4^{°}$.

The expression to calculate $\Delta H_4^{°}$ is as follows:

$\Delta H_4^{°}=\Delta H_2^{°}+\Delta H_3^{°}$ (9)

Substitute $32.912\text{kJ}$ for $\Delta H_2^{°}$ and $-139.5\text{kJ}$ for $\Delta H_3^{°}$ in the equation (9).

$\begin{array}{c}\Delta H_4^{°}=\left(32.912\text{kJ}\right)+\left(-139.5\text{kJ}\right)\\ =-106.588\text{kJ}\\ \approx -106.6\text{kJ}\end{array}$

Conclusion

$\Delta H_{\text{rxn}}^{\text{°}}$ for the fourth reaction is $-106.6\text{kJ}$.