Question

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

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

Step-by-step solution

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.

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\(K\):

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

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

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

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.