Questions: A reaction vessel initially contains 0.456 atm H2, 0.539 atm S2, and 5.18 atm H2S. After equilibrium is established, the pressure of H2 is 0.833 atm. Calculate Kp. Give your answer to 3 significant figures. 2 H2(g) + S2(g) ↔ 2 H2S(g)

Solution Steps

The balanced chemical equation is: \[ 2 \mathrm{H}_{2}(\mathrm{~g}) + \mathrm{S}_{2}(\mathrm{~g}) \leftrightarrow 2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \]

Initial pressures:

• $\mathrm{H}_{2}$: $0.456 \mathrm{~atm}$
• $\mathrm{S}_{2}$: $0.539 \mathrm{~atm}$
• $\mathrm{H}_{2} \mathrm{~S}$: $5.18 \mathrm{~atm}$

Let the change in pressure of $\mathrm{H}_{2}$ be $x$. Since the reaction consumes $2 \mathrm{H}_{2}$ for every $1 \mathrm{S}_{2}$ and produces $2 \mathrm{H}_{2} \mathrm{~S}$, the changes in pressure are:

• $\mathrm{H}_{2}$: $-2x$
• $\mathrm{S}_{2}$: $-x$
• $\mathrm{H}_{2} \mathrm{~S}$: $+2x$

Given that the equilibrium pressure of $\mathrm{H}_{2}$ is $0.833 \mathrm{~atm}$, we can write: \[ 0.456 - 2x = 0.833 \]

Solving the equation for $x$: \[ 0.456 - 2x = 0.833 \implies -2x = 0.833 - 0.456 \implies -2x = 0.377 \implies x = -0.1885 \]

Using $x = -0.1885$:

• $\mathrm{H}_{2}$: $0.456 - 2(-0.1885) = 0.456 + 0.377 = 0.833 \mathrm{~atm}$ (matches given equilibrium pressure)
• $\mathrm{S}_{2}$: $0.539 - (-0.1885) = 0.539 + 0.1885 = 0.7275 \mathrm{~atm}$
• $\mathrm{H}_{2} \mathrm{~S}$: $5.18 + 2(-0.1885) = 5.18 - 0.377 = 4.803 \mathrm{~atm}$

The equilibrium constant \( K_{\mathrm{p}} \) for the reaction is given by: \[ K_{\mathrm{p}} = \frac{(\mathrm{P}_{\mathrm{H}_{2} \mathrm{~S}})^2}{(\mathrm{P}_{\mathrm{H}_{2}})^2 (\mathrm{P}_{\mathrm{S}_{2}})} \]

Substitute the equilibrium pressures: \[ K_{\mathrm{p}} = \frac{(4.803)^2}{(0.833)^2 (0.7275)} \]

Perform the calculation: \[ K_{\mathrm{p}} = \frac{23.0736}{0.6941 \times 0.7275} = \frac{23.0736}{0.5051} \approx 45.69 \]

Final Answer