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.

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.
Transcript text: Consider the following equilibrium at 333 K : \[ \begin{array}{r} \mathrm{H} 2 \mathrm{~S}(\mathrm{~g})+\mathrm{I} 2(\mathrm{~s}) \rightleftharpoons \mathrm{S}(\mathrm{~s})+2 \mathrm{HI}(\mathrm{~g}) \\ K p=1.34 \times 10^{\wedge}-5 \end{array} \] H 2 S at a partial pressure of 786.2 mmHg and 1.68 g of $\mathrm{I} 2(\mathrm{mM}=253.80 \mathrm{~g} \mathrm{~mol}-1)$ 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. INSTRUCTIONS: Give your answer to 3 significant digits in scientific notation (example: $1.23 \times 10^{\wedge}-4$ would be typed in as 1.23E-4) in the MS Form Formatting may be lost in the question above - see the slide shown in class for correct formatting Enter your answer
failed

Solution

failed
failed

Solution Steps

Step 1: Write the equilibrium expression

The equilibrium constant Kp K_p for the reaction is given by: Kp=PHI2PH2S K_p = \frac{P_{\mathrm{HI}}^2}{P_{\mathrm{H_2S}}}

Step 2: Convert initial conditions to appropriate units

Given:

  • Partial pressure of H2S \mathrm{H_2S} is 786.2 mmHg.
  • Mass of I2 \mathrm{I_2} is 1.68 g.
  • Molar mass of I2 \mathrm{I_2} is 253.80 g/mol.
  • Volume of the flask is 1.00 L.

Calculate the initial moles of I2 \mathrm{I_2} : moles of I2=1.68g253.80g/mol=0.00662mol \text{moles of } \mathrm{I_2} = \frac{1.68 \, \text{g}}{253.80 \, \text{g/mol}} = 0.00662 \, \text{mol}

Step 3: Set up the ICE table

Initial conditions: H2SI2HIInitial786.2mmHg0.00662mol0Changexx+2xEquilibrium786.2x0.00662x2x \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}

Step 4: Apply the equilibrium expression

At equilibrium: Kp=(2x)2786.2x K_p = \frac{(2x)^2}{786.2 - x}

Given Kp=1.34×105 K_p = 1.34 \times 10^{-5} : 1.34×105=4x2786.2x 1.34 \times 10^{-5} = \frac{4x^2}{786.2 - x}

Step 5: Solve for x x

Assume x x is small compared to 786.2: 1.34×1054x2786.2 1.34 \times 10^{-5} \approx \frac{4x^2}{786.2}

Solve for x x : 4x2=1.34×105×786.2 4x^2 = 1.34 \times 10^{-5} \times 786.2 4x2=0.01053 4x^2 = 0.01053 x2=0.0026325 x^2 = 0.0026325 x=0.00263250.0513 x = \sqrt{0.0026325} \approx 0.0513

Step 6: Calculate the partial pressure of HI

The partial pressure of HI \mathrm{HI} at equilibrium is: PHI=2x=2×0.0513=0.1026atm P_{\mathrm{HI}} = 2x = 2 \times 0.0513 = 0.1026 \, \text{atm}

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

Step 7: Express the final answer in scientific notation

To 3 significant digits: PHI=7.80×101mmHg P_{\mathrm{HI}} = 7.80 \times 10^1 \, \text{mmHg}

Thus, the partial pressure of HI at equilibrium is 7.80×101 7.80 \times 10^1 mmHg.

Final Answer

7.80E1\boxed{7.80E1}

Was this solution helpful?
failed
Unhelpful
failed
Helpful