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Chapter 16: Q16.4-51E (page 911)

In the laboratory, hydrogen chloride \({\bf{(HCl(g))}}\) and ammonia \(\left( {{\bf{N}}{{\bf{H}}_{\bf{3}}}{\bf{(g)}}} \right)\)often escape from bottles of their solutions and react to form the ammonium chloride\(\left( {{\bf{N}}{{\bf{H}}_{\bf{4}}}{\bf{Cl(s)}}} \right)\), the white glaze often seen on glassware. Assuming that the number of moles of each gas that escapes into the room is the same, what is the maximum partial pressure of \({\bf{HCl}}\) and \({\bf{N}}{{\bf{H}}_{\bf{3}}}\)in the laboratory at room temperature? (Hint: The partial pressures will be equal and are at their maximum value when at equilibrium.)

Short Answer

Expert verified

Therefore the maximum partial pressures of both ammonia and hydrogen chloride are \(1.04 \times {10^{ - 8}}\)atm.

Step by step solution

01

Determining the partial pressure.

Each gas in a gas mixture contributes to the total pressure of the mixture. The partial pressure is the result of this contribution. The partial pressure of a gas is the pressure it would have if it were in the same volume and temperature as itself.

According to Dalton's law, the total pressure of an ideal gas mixture is the sum of the partial pressures of each individual gas.

02

Calculate the equilibrium constant.

The reaction in the problem is:

\({\rm{N}}{{\rm{H}}_3}(g) + {\rm{HCl}}(g) \to {\rm{N}}{{\rm{H}}_4}{\rm{Cl}}(s)\)

The equilibrium constant for this reaction is:

\(K = \frac{{x\left( {{\rm{N}}{{\rm{H}}_4}{\rm{Cl}}} \right)}}{{p\left( {{\rm{N}}{{\rm{H}}_3}} \right) \cdot p({\rm{HCl}})}}\)

And since\(x\left( {{\rm{N}}{{\rm{H}}_4}{\rm{Cl}}} \right) = 1\),

\(K = \frac{1}{{p\left( {{\rm{N}}{{\rm{H}}_3}} \right) \cdot p({\rm{HCl}})}}\)

Because the partial pressures of both gasses are equal, we can write that\(p\left( {{\rm{N}}{{\rm{H}}_3}} \right) = p({\rm{HCl}}) = x\), and therefore

\(K = \frac{1}{{{x^2}}}\)

So, that means that \(x\)equals to:

\(x = \sqrt {\frac{1}{K}} \)

Now we need to calculate the equilibrium constant, \(K\). We can do this using the formula

\(\Delta G = - RT\ln K\)

that is

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

We can calculate the \(\Delta G\)using the formula

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

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

\(\Delta G = {G_f}\left( {{\rm{N}}{{\rm{H}}_4}{\rm{Cl}},s} \right) - {G_f}\left( {{\rm{N}}{{\rm{H}}_3},g} \right) - {G_f}({\rm{HCl}},g)\)

\(\Delta G = ( - 202.87 - ( - 16.5) - ( - 95.299))\frac{{{\rm{kJ}}}}{{{\rm{mol}}}}\)

\(\Delta G = - 91.1\frac{{{\rm{kJ}}}}{{{\rm{mol}}}}\)

\(\Delta G = - 91100\frac{{\rm{J}}}{{{\rm{mol}}}}\)

We can convert the energy change from \(\frac{{kJ}}{{mol}}\)to \(\frac{J}{{mol}}\)by multiplying it by\(1000(1\;{\rm{kJ}} = 1000\;{\rm{J}})\). The temperature is room temperature, \(25{\rm{C}}\)or \(298\;{\rm{K}}\). We can convert the temperature in degrees Celsius to Kelvins by adding 273 to the temperature in degrees Celsius. Now we can calculate the equilibrium constant\(K\):

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

\(K = {e^{ - \frac{{ - 91100{\rm{Jmo}}{{\rm{l}}^{ - 1}}}}{{8.314{\rm{J}}{{\rm{K}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} \cdot 298\;{\rm{K}}}}}}\)

\(K = 9.31 \cdot {10^{15}}\)

03

Calculating the partial pressure.

Now we calculate the partial pressure

\(x = \sqrt {\frac{1}{K}} \)

\(x = \sqrt {\frac{1}{{9.31 \cdot {{10}^{15}}}}} \)

\(x = 1.04 \cdot {10^{ - 8}}\)

Since \(x = p\left( {{\rm{N}}{{\rm{H}}_3}} \right) = p({\rm{HCl}})\) and their partial pressures are at maximum value when in equilibrium, the maximum partial pressures of both ammonia and hydrogen chloride are\(1.04 \times {10^{ - 8}}\;{\rm{atm}}\).

Therefore the maximum partial pressures of both ammonia and hydrogen chloride are \(1.04 \times {10^{ - 8}}\)atm.

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Most popular questions from this chapter

Predict the sign of the enthalpy change for the following processes. Give a reason for your prediction.

(a) \({\bf{NaN}}{{\bf{O}}_{\bf{3}}}{\bf{(s)}} \to {\bf{N}}{{\bf{a}}^{\bf{ + }}}{\bf{(aq) + N}}{{\bf{O}}_{\bf{3}}}^{\bf{ - }}{\bf{(aq)}}\)

(b) the freezing of liquid water

(c) \({\bf{C}}{{\bf{O}}_{\bf{2}}}{\bf{(s)}} \to {\bf{C}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

(d) \({\bf{CaCO(s)}} \to {\bf{CaO(s) + C}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

Calculate the standard entropy change for the following process:

\({{\bf{H}}_{\bf{2}}}{\bf{(g) + }}{{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{4}}}{\bf{(g)}} \to {{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{6}}}{\bf{(g)}}\)

Popular chemical hand warmers generate heat by the air-oxidation of iron:\({\bf{4Fe(s) + 3}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2F}}{{\bf{e}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(s)}}\).How does the spontaneity of this process depend upon temperature?

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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