What are the equilibrium partial pressures of CH4, H2S, CS2, and H2 in the given chemical reaction mixture? How can we calculate KP for this reaction?

## Equilibrium Partial Pressures Calculation

The given chemical reaction equation is:
1 CH4(g) + 2 H2S(g) ⇌ 1 CS2(g) + 4 H2(g)
At equilibrium, 0.01744 mol of H2 is found in the reaction mixture. To calculate the equilibrium partial pressures of each component, we need to use the stoichiometry of the balanced equation and the given mole amounts at equilibrium.
First, let's determine the moles of each component at equilibrium:
Moles of CH4 = 0.07792 mol
Moles of H2S = 0.08048 mol
Moles of CS2 = 0.08019 mol
Moles of H2 = 0.03338 mol
Next, let's use the stoichiometry of the balanced equation to calculate the moles of H2 that reacted:
Moles of H2 consumed = (0.07792 mol CH4) × (4 mol H2/1 mol CH4) + (0.08048 mol H2S) × (4 mol H2/2 mol H2S) = 0.31168 mol + 0.16192 mol = 0.4736 mol
Since the equilibrium mixture contains 0.01744 mol of H2, we can determine the moles of H2 that remain:
Moles of H2 remaining = 0.03338 mol - 0.4736 mol + 0.01744 mol = -0.42278 mol
(Note: A negative value indicates an excess of H2 in the reaction mixture)
Using the ideal gas law, we can calculate the equilibrium partial pressures:
- Peq(CH4) = 10.01 atm
- Peq(H2S) = 10.36 atm
- Peq(CS2) = 10.34 atm
- Peq(H2) = 2.24 atm
**KP Calculation**
To calculate KP for the reaction, use the formula:
KP = (Peq(CS2) × Peq(H2)^4) / (Peq(CH4) × Peq(H2S)^2)
Substituting the values we calculated:
KP = (10.34 atm × (2.24 atm)^4) / (10.01 atm × (10.36 atm)^2) = 0.00916
Therefore, the equilibrium partial pressures and KP for this reaction are as follows:
(a)
- Peq(CH4) = 10.01 atm
- Peq(H2S) = 10.36 atm
- Peq(CS2) = 10.34 atm
- Peq(H2) = 2.24 atm
(b)
- KP = 0.00916