If the equilibrium constant K equals .05, and the reaction C3H8=C3H6 + H2 (all gases), starts with pure C3H8 and is allowed to reach equilibrium at 200 K at a total pressure of 2 bar. Calculate the partial pressure of H2 at equilbirum.
Is that Kp in bar?
........C3H8 ==> C3H6 + H2
Total P = 2 = pC3H6 + pH2 + pC3H8
Solve for pC3H8 = 2-p-p = 2-2p
Kp = p*p/(2-2p)
Solve for p = H2 = C3H6
The K is just the equilibrium constant, not Kp. Do you have to convert the K to Kp?
To calculate the partial pressure of H2 at equilibrium, we need to use the equilibrium expression and the given equilibrium constant value. Here's how you can solve it step by step:
Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
C3H8 → C3H6 + H2
Step 2: Write the equilibrium expression
The equilibrium expression for the given reaction is:
K = [C3H6] * [H2] / [C3H8]
Since all the gases are involved, we don't need to consider their concentrations, but the ratio of their partial pressures.
Step 3: Determine the initial and equilibrium partial pressures
Since pure C3H8 is initially present, its partial pressure is 2 bar (given in the question).
Let's assume the partial pressure of H2 at equilibrium is 'x'.
So, the equilibrium partial pressures are:
[C3H8] = 2 bar
[C3H6] = 0 (as it is not initially present)
[H2] = x
Step 4: Substitute the values into the equilibrium expression
We substitute the given values into the equilibrium expression and solve for 'x':
K = [C3H6] * [H2] / [C3H8]
0.05 = 0 * x / 2
0.05 = 0
Since [C3H6] is zero, the value of 'x' becomes irrelevant in this case.
Therefore, the partial pressure of H2 at equilibrium is zero.
Note: This result indicates that the reaction predominantly proceeds in the forward direction, resulting in the consumption of all available C3H8 and formation of C3H6. The production of H2 is negligible or nonexistent under these conditions.