Question

Consider the decomposition of ammonium hydrogen sulfide: $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ In a sealed flask at \(25^{\circ} \mathrm{C}\) are \(10.0 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{HS}\), ammonia with a partial pressure of \(0.692 \mathrm{~atm},\) and \(\mathrm{H}_{2} \mathrm{~S}\) with a partial pressure of \(0.0532 \mathrm{~atm}\). When equilibrium is established, it is found that the partial pressure of ammonia has increased by 12.4\%. Calculate \(K\) for the decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) at \(25^{\circ} \mathrm{C}\).

Short answer

Question: Calculate the equilibrium constant, K, for the decomposition of ammonium hydrogen sulfide (NH4HS) into ammonia (NH3) and hydrogen sulfide (H2S) at 25°C, given that the initial mass of NH4HS is 10.0 g and the initial partial pressures of NH3 and H2S are 0.692 atm and 0.0532 atm, respectively. The partial pressure of ammonia increased by 12.4% at equilibrium. Answer: The equilibrium constant, K, for the decomposition of NH4HS at 25°C is approximately 0.416.

Step-by-step solution

Calculate the number of moles of NH4HS, NH3, and H2S

We first convert the masses of NH4HS into moles using the molar mass. The molar mass of NH4HS is 53.1 g/mol (14 + 1*4 + 1 + 32 + 16). From the ideal gas law, we can also convert the partial pressures of NH3 and H2S into moles, using the total volume V (V = nRT/P): $$ n_{\mathrm{NH}_{4}\mathrm{HS}} = \frac{10.0 \mathrm{~g}}{53.1 \mathrm{~g/mol}} = 0.188 \mathrm{~mol}\\ n_{\mathrm{NH}_{3}} = \frac{P_{\mathrm{NH}_{3}}V}{RT} = \frac{0.692 \mathrm{~atm} \times V}{(0.0821 \mathrm{~L \cdot atm/mol \cdot K}) (298 \mathrm{~K})} = \frac{0.692V}{24.51} \mathrm{~mol}\\ n_{\mathrm{H}_{2}\mathrm{S}} = \frac{P_{\mathrm{H}_{2}\mathrm{S}}V}{RT} = \frac{0.0532 \mathrm{~atm} \times V}{(0.0821 \mathrm{~L \cdot atm/mol \cdot K}) (298 \mathrm{~K})} = \frac{0.0532V}{24.51} \mathrm{~mol} $$

Set up the ICE table

Next, we set up an ICE (Initial-Change-Equilibrium) table for the reaction. - Initial moles: [NH4HS]=0.188 mol, [NH3]=0.692V/24.51 mol, [H2S]=0.0532V/24.51 mol - Change in moles: As NH4HS decomposes, NH3 increases by 12.4%, so Δ[NH3]=0.124[initial NH3] = 0.124(0.692V/24.51) = 0.0346V/24.51 mol. Since the stoichiometry of the reaction is 1:1:1, the change in moles for NH4HS and H2S are the same as for NH3. - Equilibrium moles: [NH4HS]=0.188 - 0.0346V/24.51 mol, [NH3]=0.692V/24.51 + 0.0346V/24.51 mol, [H2S]=0.0532V/24.51 + 0.0346V/24.51 mol

Calculate the value of K

Now, we apply the definition of the equilibrium constant K: $$ K = \frac{[\mathrm{NH}_{3}][\mathrm{H}_{2} \mathrm{S}]}{[\mathrm{NH}_{4} \mathrm{HS}]} $$ We already have the equilibrium moles of each species, so we can substitute them into the expression: $$ K = \frac{[\frac{0.692V}{24.51} + \frac{0.0346V}{24.51}] [\frac{0.0532V}{24.51} + \frac{0.0346V}{24.51}]}{[\frac{0.188V}{24.51} - \frac{0.0346V}{24.51}]} $$ Notice that all terms have V/24.51, and we can cancel them, therefore: $$ K = \frac{(0.692 + 0.0346)(0.0532 + 0.0346)}{0.188 - 0.0346} $$ Now, we calculate the value of K: $$ K =\frac{(0.7266)(0.0878)}{0.1534} = 0.416 $$

Present the final answer

The equilibrium constant of the decomposition of NH4HS at 25 °C is approximately 0.416.