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Chapter 16: Q16.4-54E (page 911)

Carbon tetrachloride, an important industrial solvent, is prepared by the chlorination of methane at \(850\;{\rm{K}}\).

\({\rm{C}}{{\rm{H}}_4}(g) + 4{\rm{C}}{{\rm{l}}_2}(g) \to {\rm{CC}}{{\rm{l}}_4}(g) + 4{\rm{HCl}}(g)\)

What is the equilibrium constant for the reaction at \(850\;{\rm{K}}\)? Would the reaction vessel need to be heated or cooled to keep the temperature of the reaction constant?

Short Answer

Expert verified

The equilibrium constant for the reaction is \(K = 7.9 \times {10^{23}}\) and the reaction vessel needs to be cooled down to keep the temperature constant.

Step by step solution

01

Definition of Equilibrium constant.

A chemical response's equilibrium constant is the value of its response quotient at chemical equilibrium, a state reached by a dynamic chemical system after a period of time has passed in which its composition shows no perceptible tendency to change.

02

Determine the equilibrium constant.

In order to calculate the equilibrium constant for the reaction, we can use the formula\(\Delta G = - RT\ln K\)

that is

\(K = {e^{ - \frac{{\Delta G}}{{RT}}}}\)

We can use the formula

\(\Delta G = {G_f}({\rm{ products }}) - {G_f}({\rm{ reactants }})\)

To calculate the\(\Delta G\). We can look up the standard formation free energy changes for each species in Appendix G in the book.

\(\begin{array}{l}\Delta G = {G_f}\left( {{\rm{CC}}{{\rm{l}}_4},g} \right) + 4{G_f}(HCl,g) - {G_f}\left( {{\rm{C}}{{\rm{H}}_4},g} \right) - 4{G_f}(C{l_2},g)\\\Delta G = ( - 58.2 + 4 \cdot ( - 95.299) - ( - 50.5) - 4 \cdot 0)\frac{{kJ}}{{mol}}\\\Delta G = - 388.9\frac{{kJ}}{{mol}}\\\Delta G = - 388900\frac{J}{{mol}}\end{array}\)

To convert the free energy change from\(\frac{{kJ}}{{mol}}\)to\(\frac{J}{{mol}}\)we multiply the energy change by\(1000(1kJ = 1000J)\). Now we can calculate the equilibrium constant:

\(K = {e^{ - \frac{{\Delta G}}{{RT}}}}\)

\(K = {e^{ - \frac{{ - 388900{\rm{Jmo}}{{\rm{l}}^{ - 1}}}}{{8.314{\rm{J}}{{\rm{K}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} \cdot 850\;{\rm{K}}}}}}\)

\(K = 7.9 \times {10^{23}}\)

03

Determine the reaction is heated or cooled to keep the reaction constant.

To determine whether the reaction vessel needs to be heated or cooled to keep the temperature constant, we have to think about is the heat released by the system or taken in by the system.

If the heat is released, the reaction vessel heats so we would have to cool it down, otherwise, the vessel would become cooler because the system absorbs heat from the environment.

The enthalpy change tells us this. To calculate the enthalpy change, we can use the formula

\(\Delta H = {H_f}({\rm{ products }}) - {H_f}({\rm{ reactants }})\)

We can look up the formation enthalpy changes for each species in the table in Appendix G in the book.

\(\begin{array}{l}\Delta H = {H_f}\left( {CC{l_4},g} \right) + 4{H_f}(HCl,g) - {H_f}\left( {C{H_4},g} \right) - 4{H_f}\left( {C{l_2},g} \right)\\\Delta H = ( - 95.7 + 4 \cdot ( - 92.307) - ( - 74.6) - 4 \cdot 0)\frac{{kJ}}{{mol}}\\\Delta H = - 390.3\frac{{kJ}}{{mol}}\end{array}\)

Since the enthalpy change for this reaction is negative, it is an exothermic reaction. Exothermic reactions release heat into the environment this means that the reaction vessel heats during the reaction and therefore needs to be cooled down to keep the temperature constant

Therefore, the equilibrium constant value is \(K = 7.9 \times {10^{23}}\)and the reaction vessel needs to be cooled down to keep the temperature constant.

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Most popular questions from this chapter

What is a spontaneous reaction?

Under what conditions is \({{\bf{N}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(g)}} \to {\bf{NO(g) + N}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\) spontaneous?

Among other things, an ideal fuel for the control thrusters of a space vehicle should decompose in a spontaneous exothermic reaction when exposed to the appropriate catalyst. Evaluate the following substances under standard state conditions as suitable candidates for fuels.

(a) Ammonia\({\bf{:}}{\rm{ }}{\bf{2N}}{{\bf{H}}_{\bf{3}}}{\bf{(g)}} \to {{\bf{N}}_{\bf{2}}}{\bf{(g) + 3}}{{\bf{H}}_{\bf{2}}}{\bf{(g)}}\)

(b) Diborane\({\bf{:}}{\rm{ }}{{\bf{B}}_{\bf{2}}}{{\bf{H}}_{\bf{6}}}{\bf{(g)}} \to {\bf{2\;B(g) + 3}}{{\bf{H}}_{\bf{2}}}{\bf{(g)}}\)

(c) Hydrazine: \({{\bf{N}}_{\bf{2}}}{{\bf{H}}_{\bf{4}}}{\bf{(g)}} \to {{\bf{N}}_{\bf{2}}}{\bf{(g) + 2}}{{\bf{H}}_{\bf{2}}}{\bf{(g)}}\)

(d) Hydrogen peroxide: \({{\bf{H}}_{\bf{2}}}{{\bf{O}}_{\bf{2}}}{\bf{(l)}} \to {{\bf{H}}_{\bf{2}}}{\bf{O(g) + }}\frac{{\bf{1}}}{{\bf{2}}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

What happens to \({\bf{\Delta G}}_{{\bf{298}}}^{\bf{^\circ }}\) (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?

(a) \({\bf{S(s) + }}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{S}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

(b) \({\bf{2S}}{{\bf{O}}_{\bf{2}}}{\bf{(g) + }}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{S}}{{\bf{O}}_{\bf{3}}}{\bf{(g)}}\)

(c) \({\bf{HgO(s)}} \to {\bf{Hg(l) + }}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

"Thermite" reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is \({\bf{F}}{{\bf{e}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(s) + 2Al(s)}} \to {\bf{A}}{{\bf{l}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(s) + 2Fe(s)}}\). Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb \({\bf{851}}{\bf{.8\;kJ/mol}}\)of heat.
See all solutions

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