Questions: Consider the following equilibrium at 333 K: H2S(g) + I2(s) ⇌ S(s) + 2 HI(g) Kp = 1.34 × 10^-5 H2S at a partial pressure of 786.2 mmHg and 1.68 g of I2 (MM = 253.80 g mol^-1) are added to an otherwise empty 1.00 L flask at 333 K and allowed to come to equilibrium. Determine the partial pressure of HI in mmHg at equilibrium.

Solution Steps

The equilibrium constant \( K_p \) for the reaction is given by: \[ K_p = \frac{P_{\mathrm{HI}}^2}{P_{\mathrm{H_2S}}} \]

Given:

• Partial pressure of \( \mathrm{H_2S} \) is 786.2 mmHg.
• Mass of \( \mathrm{I_2} \) is 1.68 g.
• Molar mass of \( \mathrm{I_2} \) is 253.80 g/mol.
• Volume of the flask is 1.00 L.

Calculate the initial moles of \( \mathrm{I_2} \): \[ \text{moles of } \mathrm{I_2} = \frac{1.68 \, \text{g}}{253.80 \, \text{g/mol}} = 0.00662 \, \text{mol} \]

Initial conditions: \[ \begin{array}{c|c|c|c} & \mathrm{H_2S} & \mathrm{I_2} & \mathrm{HI} \\ \hline \text{Initial} & 786.2 \, \text{mmHg} & 0.00662 \, \text{mol} & 0 \\ \text{Change} & -x & -x & +2x \\ \text{Equilibrium} & 786.2 - x & 0.00662 - x & 2x \\ \end{array} \]

At equilibrium: \[ K_p = \frac{(2x)^2}{786.2 - x} \]

Given \( K_p = 1.34 \times 10^{-5} \): \[ 1.34 \times 10^{-5} = \frac{4x^2}{786.2 - x} \]

Assume \( x \) is small compared to 786.2: \[ 1.34 \times 10^{-5} \approx \frac{4x^2}{786.2} \]

Solve for \( x \): \[ 4x^2 = 1.34 \times 10^{-5} \times 786.2 \] \[ 4x^2 = 0.01053 \] \[ x^2 = 0.0026325 \] \[ x = \sqrt{0.0026325} \approx 0.0513 \]

The partial pressure of \( \mathrm{HI} \) at equilibrium is: \[ P_{\mathrm{HI}} = 2x = 2 \times 0.0513 = 0.1026 \, \text{atm} \]

Convert to mmHg: \[ P_{\mathrm{HI}} = 0.1026 \, \text{atm} \times 760 \, \text{mmHg/atm} = 77.976 \, \text{mmHg} \]

To 3 significant digits: \[ P_{\mathrm{HI}} = 7.80 \times 10^1 \, \text{mmHg} \]

Thus, the partial pressure of HI at equilibrium is \( 7.80 \times 10^1 \) mmHg.

Final Answer