Chapter 6, Problem 6.78P

Silicon tetrachloride is produced annually on the multikiloton scale and used in making transistor-grade silicon. It can br produced directly from the elements (reaction 1) or, more cheaply, by heating sand and graphite with chlorine gas (reaction 2). If water is present in reaction 2, some tetrachloride may be lost in an unwante side reaction (reaction 3):

1. $\text{S}i(s)+2\text{C}{\text{l}}_2(g)\to \text{S}i\text{C}{\text{l}}_4(g)$ $\text{S}i{\text{O}}_2(s)+2\text{C}(graphit\text{e})+{\text{ 2Cl}}_{\text{2}}(g)\to \text{S}i\text{C}{\text{l}}_4(g)+2\text{C}\text{O}(g)$ $\text{S}i\text{C}{\text{l}}_4(g)+2{\text{H}}_2\text{O}(g)\to \text{S}i{\text{O}}_2(s)+4\text{H}\text{C}\text{l}(g)$          $\text{Δ}{\text{H}}_{rxn}^o=-139.5\text{kJ}$ Use reaction 3 to calculate the standard enthalpies of reaction of reactions 1 and 2. (b)What is the standard enthalpy of reaction for a fourth reaction that is the sum of reactions 2 and 3?

Interpretation Introduction

Interpretation:Thestandard enthalpy of reaction needs to be determined.

Concept Introduction: For a general reaction as follows:

  $A+B\to C\text{}+D$

The change in enthalpy of reaction can be calculated as follows:

  $\Delta {\text{H}}^o{}_{rxn}\text{=}{\displaystyle \sum \Delta {\text{H}}_f^o\left(\text{product}\right)-}{\displaystyle \sum \Delta {\text{H}}_f^o\left(\text{reactant }\right)}$

Answer to Problem 6.78P

  $\begin{array}{l}(\Delta H^0{}_{rxn})\text{for}\text{equation}\text{(1)}\text{=}\text{-657}\text{.0}\text{kJ}\\ (\Delta H^0{}_{rxn})\text{for}\text{equation}\text{(2)}\text{=}\text{32}\text{.9}\text{kJ}\end{array}$

Explanation of Solution

The chemical equations are as follows:

  $\text{Si}\text{(s)}\text{+}{\text{2Cl}}_{\text{2}}\text{(g)}\to {\text{SiCl}}_{\text{4}}\text{(g)}$..... (1)

  ${\text{SiO}}_{\text{2}}\text{(s)}\text{+}\text{2C}\text{(graphite)}\text{+}{\text{2Cl}}_{\text{2}}\text{(g}\text{}\text{)}\to {\text{SiCl}}_{\text{4}}\text{(g)}\text{+}\text{2CO}\text{(g)}$ ..... (2)

The thermochemical equation is as follows:

  ${\text{SiCl}}_{\text{4}}\text{(g)}\text{+}{\text{2H}}_{\text{2}}\text{O}\text{(g)}\to {\text{SiO}}_{\text{2}}\text{(s)}\text{+}\text{4HCl}\text{(g)}{\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{=}\text{-139}\text{.5}\text{kJ}$ ..... (3)

  $\Delta H^0{}_{rxn}=\sum m\Delta H^0{}_f(\text{products})-\sum n\Delta H^0{}_f(\text{reactants})$

Here, m and n are moles of products and reactants.

  $\begin{array}{c}\Delta H^0{}_{rxn}=\left[1\text{mol}\Delta H^0{}_f({\text{SiO}}_2\text{(s)})+4\text{mol}\Delta H^0{}_f(HCl\text{(g}))\right]-\left[1\text{mol}\Delta H^0{}_f({\text{SiCl}}_{\text{2}}\text{(g}))+2\text{mol}\Delta H^0{}_f(H_2\text{O}\text{(g)})\right]\\ =\left[1\cancel{\text{mol}}\times \left(-910.9\frac{\text{kJ}}{\cancel{\text{mol}}}\right)+4\cancel{\text{mol}}\times \left(-92.31\frac{\text{kJ}}{\cancel{\text{mol}}}\right)\right]-\left[1\text{mol}\Delta H^0{}_f({\text{SiCl}}_{\text{2}}\text{(g})+2\text{mol}\times \left(-241.8\frac{\text{kJ}}{\cancel{\text{mol}}}\right)\right]\\ -1\text{39}\text{.5}\text{kJ}\text{=}-\text{910}\text{.9}\text{kJ}-\text{369}\text{.24}\text{kJ}-\text{1}\text{mol}\Delta H^0{}_f{\text{(SiCl}}_{\text{4}}\text{(g)}\text{+}\text{483}\text{.6}\text{kJ}\end{array}$

  $\begin{array}{c}1\text{mol}\Delta H^0{}_f{\text{(SiCl}}_{\text{4}}\text{(g))}=-\text{910}\text{.9}\text{kJ}-\text{369}\text{.24}\text{kJ}\text{+}\text{483}\text{.6}\text{kJ}+1\text{39}\text{.5}\text{kJ}\\ \Delta H^0{}_f{\text{(SiCl}}_{\text{4}}\text{(g))=}\frac{-\text{657}\text{.0}\text{kJ}}{\text{1}\text{mol}}\\ \text{=}-\text{657}\text{.0}\text{kJ/mol}\end{array}$

Therefore, $\Delta H^0{}_{rxn}$ of (1) is $-\text{657}\text{.0}\text{kJ/mol}$

  $\begin{array}{c}\Delta H^0{}_{rxn}=\left[1\text{mol}\Delta H^0{}_f({\text{SiCl}}_{\text{4}}\text{(g)})+2\text{mol}\Delta H^0{}_f(\text{CO(g}))\right]-\left[\begin{array}{l}1\text{mol}\Delta H^0{}_f({\text{SiO}}_{\text{2}}\text{(g}))\\ +2\text{mol}\Delta H^0{}_f(\text{C(graphite)})\end{array}\right]\\ =\left[1\cancel{\text{mol}}\times \left(-657.0\frac{\text{kJ}}{\cancel{\text{mol}}}\right)+2\cancel{\text{mol}}\times \left(-110.5\frac{\text{kJ}}{\cancel{\text{mol}}}\right)\right]-\left[\begin{array}{l}1\cancel{\text{mol}}\times \left(-910.9\frac{\text{kJ}}{\cancel{\text{mol}}}\right)\\ +2\text{mol}\times 0+\text{2}\text{mol}\text{×}\text{0}\end{array}\right]\\ \text{=}-\text{657}\text{.0}\text{kJ}-\text{221}\text{.0}\text{kJ}\text{+}\text{910}\text{.9}\text{kJ}\\ \text{=}\text{32}\text{.9}\text{kJ}\end{array}$

Thus, $\Delta H^{0}rxn$ of (2) is $\text{32}\text{.9}\text{kJ}$ .

Interpretation Introduction

Interpretation:The standard enthalpy of reaction for the fourth reaction needs to be determined.

Concept Introduction : According to Hess’s law, the change of enthalpy in a process is the summation of enthalpy changes of individual steps of the reaction.

  $\Delta H_{\text{overall}}=\Delta H_1+\Delta H_2+\mathrm{.......}+\Delta H_n$

Answer to Problem 6.78P

The heat of reaction for the new reaction is $\text{-106}\text{.6}\text{kJ}$ .

Explanation of Solution

The thermochemical equations are as follows:

  ${\text{SiO}}_{\text{2}}\text{(s)}\text{+}\text{2C}\text{(graphite)}\text{+}{\text{2Cl}}_{\text{2}}\text{(g}\text{}\text{)}\to {\text{SiCl}}_{\text{4}}\text{(g)}\text{+}\text{2CO}\text{(g)}$ …… (2)

  ${\text{SiCl}}_{\text{4}}\text{(g)}\text{+}{\text{2H}}_{\text{2}}\text{O}\text{(g)}\to {\text{SiO}}_{\text{2}}\text{(s)}\text{+}\text{4HCl}\text{(g)}{\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{=}\text{-139}\text{.5}\text{kJ}$ …… (3)

Now add (2) and (3).

  $\begin{array}{l}\cancel{{\text{SiO}}_{\text{2}}\text{(s)}}\text{+}\text{2C}\text{(graphite)}\text{+}{\text{2Cl}}_{\text{2}}\text{(g}\text{}\text{)}\to \cancel{{\text{SiCl}}_{\text{4}}\text{(g)}}\text{+}\text{2CO}\text{(g)}{\text{. ΔH}}^{\text{0}}{}_{\text{rxn}}=32.9kJ\\ \cancel{{\text{SiCl}}_{\text{4}}\text{(g)}}\text{+}{\text{2H}}_{\text{2}}\text{O}\text{(g)}\to \cancel{{\text{SiO}}_{\text{2}}\text{(s)}}\text{+}\text{4HCl}\text{(g)}{\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{=}\text{-139}\text{.5}\text{kJ}\end{array}$

New reaction is,

  $\begin{array}{l}\text{2C}\text{(graphite)}+{\text{2Cl}}_{\text{2}}\text{(g}\text{})+{\text{2H}}_2\text{O}\text{(g)}\to \text{4HCl}\text{(g)}+2\text{CO}\text{(g)}{\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{=}\text{-106}\text{.6}\text{kJ}\\ \end{array}$

Thus, the heat of the reaction for a new reaction is $\text{-106}\text{.6}\text{kJ}$ .