At a certain temperature, the Kp for the decomposition of H2S is 0.734.Initially, only H2S is present at a pressure of 0.102 atm in a closed container. What is the total pressure in the container at equilibrium?
Is this the equation? You should have given it.
.........H2S ==> H2 + S
I......0.102......0...0
C........-p.......p...p
E......0.102-p....p...p
Substitut the E line into the Kp expression and solve for p, then evaluate 0.102-p. Finally, add the partial pressures and the sum is the total pressure.
To find the total pressure at equilibrium, we need to use the given value of Kp and the initial pressure of H2S.
Let's denote the equilibrium pressure of H2S as "x" atm. According to the balanced chemical equation for the decomposition of H2S:
2H2S(g) ⇌ 2H2(g) + S2(g)
We can write the expression for Kp:
Kp = (P(H2))^2 * P(S2) / (P(H2S))^2
Given that Kp = 0.734, and the initial pressure of H2S is 0.102 atm, the equation becomes:
0.734 = (P(H2))^2 * P(S2) / (0.102)^2
Since the stoichiometric coefficients for H2(g) and S2(g) are both 2, their partial pressures will be equal at equilibrium. Therefore, we can substitute P(H2) = P(S2) = x into the equation:
0.734 = (x)^2 * x^2 / (0.102)^2
Rearranging the equation, we get:
0.734 * (0.102)^2 = x^4
Simplifying, we find:
0.007574196 = x^4
Taking the fourth root of both sides, we get:
x = ∛0.007574196
Calculating this, we find that x ≈ 0.282 atm.
Therefore, the total pressure in the container at equilibrium is the sum of the equilibrium pressures of all the gases:
Total pressure = P(H2S) + P(H2) + P(S2)
= 0.102 atm + 0.282 atm + 0.282 atm
= 0.666 atm
Thus, the total pressure in the container at equilibrium is approximately 0.666 atm.