A mixture of 1.00 g H2 and 1.06 g H2S in a 0.500−L flask comes to equilibrium at 1670 K: 2H2(g)+S2(g)⇌2H2S. The equilibrium amount of S2(g) found is 8.00×10−6mol.

Determine the value of KP at 1670K
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To determine the equilibrium constant \( K_P \) for the reaction at 1670 K, we first need to analyze the given information and apply the equilibrium expression for the reaction. The balanced equation is:

Reaction Overview

2H₂(g) + S₂(g) ⇌ 2H₂S(g)

Step 1: Calculate Moles of Reactants

We start by calculating the number of moles of H₂ and H₂S in the mixture.

• Molar mass of H₂ = 2.02 g/mol
• Moles of H₂ = 1.00 g / 2.02 g/mol ≈ 0.495 mol

• Molar mass of H₂S = 34.08 g/mol
• Moles of H₂S = 1.06 g / 34.08 g/mol ≈ 0.031 mol

Step 2: Initial Concentrations

Next, we calculate the initial concentrations of H₂ and H₂S in the 0.500 L flask.

• Initial concentration of H₂ = 0.495 mol / 0.500 L = 0.990 M
• Initial concentration of H₂S = 0.031 mol / 0.500 L = 0.062 M

Step 3: Change in Concentrations at Equilibrium

At equilibrium, we know that the amount of S₂ is 8.00 × 10^-6 mol. We can find the concentration of S₂:

• Concentration of S₂ = 8.00 × 10^-6 mol / 0.500 L = 1.60 × 10^-5 M

From the stoichiometry of the reaction, for every 2 moles of H₂ consumed, 1 mole of S₂ is produced. Therefore, the change in concentration of H₂ is:

• Change in H₂ = 2 × (1.60 × 10^-5 M) = 3.20 × 10^-5 M

Step 4: Equilibrium Concentrations

Now we can find the equilibrium concentrations:

• Equilibrium concentration of H₂ = 0.990 M - 3.20 × 10^-5 M ≈ 0.990 M
• Equilibrium concentration of H₂S = 0.062 M + 1.60 × 10^-5 M ≈ 0.062 M

Step 5: Equilibrium Constant Expression

The equilibrium constant \( K_P \) for the reaction is given by:

K_P = \frac{(P_H2S)^2}{(P_H2)^2 (P_S2)}

Step 6: Calculate Partial Pressures

To find the partial pressures, we can use the ideal gas law, \( P = C \cdot RT \), where \( R = 0.0821 \, \text{L·atm/(K·mol)} \) and \( T = 1670 \, K \).

• Partial pressure of H₂ = 0.990 M × 0.0821 L·atm/(K·mol) × 1670 K ≈ 123.2 atm
• Partial pressure of H₂S = 0.062 M × 0.0821 L·atm/(K·mol) × 1670 K ≈ 8.54 atm
• Partial pressure of S₂ = 1.60 × 10^-5 M × 0.0821 L·atm/(K·mol) × 1670 K ≈ 2.21 × 10^-3 atm

Step 7: Plugging into the Equilibrium Expression

Now we can substitute these values into the \( K_P \) expression:

K_P = \frac{(8.54)^2}{(123.2)^2 \cdot (2.21 × 10^{-3})}

Calculating this gives:

• Numerator = (8.54)^2 ≈ 73.0
• Denominator = (123.2)^2 × (2.21 × 10^-3) ≈ 3.0 × 10³

Thus,

K_P ≈ \frac{73.0}{3.0 × 10³} ≈ 0.0243

Final Result

The equilibrium constant \( K_P \) at 1670 K is approximately 0.0243. This value indicates the extent to which the reaction favors the formation of products at this temperature.