A vessel initially contains pure A at a pressure of 1.0 bar. The total pressure of all gases at equilibrium is 1.5 bar. What is K for this reaction?
A (g) <--> B (g) + C (g)
To find the equilibrium constant (K) for this reaction, you need to know the partial pressures of the reactants and products at equilibrium.
Let's assume that at equilibrium, the partial pressure of A (PA) is x bar, the partial pressure of B (PB) is y bar, and the partial pressure of C (PC) is z bar.
At the beginning of the reaction, the vessel contains only pure A at a pressure of 1.0 bar, so PA = 1.0 bar.
According to the balanced chemical equation, the stoichiometric coefficients for A, B, and C are 1, 1, and 1 respectively.
At equilibrium, the total pressure of all gases is given as 1.5 bar. So, we have the equation:
PA + PB + PC = 1.5 bar
Substituting the values we know, we get:
1.0 bar + y bar + z bar = 1.5 bar
Now, we need to express the equilibrium concentrations in terms of partial pressures to derive the expression for the equilibrium constant (K).
Since the stoichiometric coefficients for A, B, and C are all 1, we can say that:
PA = x bar
PB = y bar
PC = z bar
Therefore, the equation becomes:
x bar + y bar + z bar = 1.5 bar
From this equation, we can deduce that x + y + z = 1.5.
The equilibrium constant expression (K) can be written as:
K = (PB * PC) / (PA)
Substituting the values we have, we get:
K = (y * z) / (x)
So, the equilibrium constant (K) for this reaction is (y * z) / (x).
........A--> B + C
I......1.0...0...0
C......-p....p...p
E.....1.0-p..p...p
Total P = 1-p+p+p = 1.5
Solve for p, evaluate 1-p, substitute into Kp expression and solve for Kp.