If the initial pressure of H2S(g) is 7.404 atm, calculate the % decomposition of H2S(g) when the reaction comes to equilibrium according to the balanced equation. The value of Kp at 1132 °C is 0.0265. The initial pressure of the reaction products is 0 atm.
Is the equation given as
H2S --> H2 + S
Set up ICE chart and substitute into Kp.
Initial:
pH2 = 0
pS = 0
pH2S = 7.404
change:
S = +x
H2 = +x
H2S = -x
equilibrium:
S = x
H2 = x
H2S = 7.404-x
To calculate the percent decomposition of H2S(g) at equilibrium, we need to use the equilibrium expression and the given values.
The balanced equation for the decomposition of H2S(g) is:
2H2S(g) ↔ 2H2(g) + S2(g)
The equilibrium expression for this reaction is:
Kp = (PH2)^2 x (PS2) / (PH2S)^2
Here, PH2, PS2, and PH2S represent the partial pressures of H2(g), S2(g), and H2S(g), respectively.
Given information:
Initial pressure of H2S(g) = 7.404 atm
Value of Kp at 1132 °C = 0.0265
Initial pressure of the reaction products (H2(g) and S2(g)) = 0 atm
Since the initial pressure of the reaction products is 0 atm, we can assume they don't contribute to the equilibrium expression. Therefore, the equilibrium expression simplifies to:
Kp = (PH2)^2 x (PS2) / (PH2S)^2
To find the percent decomposition, we need to determine the partial pressure of H2S(g) at equilibrium. Let's assume it is 'x' atm.
Using the given information, we can set up the following equation:
0.0265 = (0)^2 x (0) / (7.404 - x)^2
Simplifying the equation gives:
0.0265 = 0 / (7.404 - x)^2
Since anything divided by 0 is undefined, the only way this equation can hold is if the numerator is also 0.
Therefore, we can conclude that H2S(g) will not decompose appreciably at this temperature and pressure. The percent decomposition would be 0%.