Consider the reaction below, where Kp = 1.00 x 10-1 at 1325 K. In an experiment
ID: 552253 • Letter: C
Question
Consider the reaction below, where Kp = 1.00 x 10-1 at 1325 K. In an experiment where P4 (g) was 155 placed into a container at 1325K, the equilibrium mixture of P4 (g) and P2 (g) has a total pressure of 1.00 atm P,(g) 2P2(g) a) Calculate the equilibrium pressures of P4(g) and P2 (g). b) Calculate the fraction of P4 (g) that has dissociated to reach equilibrium c) If the volume of the container containing the equilibrium mixture of P4 (g) and P2 (g) decreases, how will the equilibrium position shift?Explanation / Answer
Here we can write the equilibrium equation as:
Kp = [P2]2 / [P4]
1 x 10-1 = [P2]2 / [P4]
[P2]2 = 0.1 x [P4]
lets say initially we had A atm of P4
P4 <---->2P2
A 0 initially
-X +2X change in partial pressures
A-X + 2X equilibrium partial pressures
total pressure = 1 atm = A-X + 2X = A+X
A = 1-X
[2X]2 = 0.1 x [A-X]
4X2 = 0.1A - 0.1X
4X2 -0.1(1-X) + 0.1X = 0
4X2 -0.1 + 0.1X + 0.1X = 0
4X2 + 0.2X - 0.1= 0
two roots are x1 = 0.135
x2 = -0.185 { neglected as the product formed is not negative}
so total pressure = 1atm = A + X
A = 1-0.135 = 0.865 atm
equilibrium pressure of P4 = A-X = 0.865 - 0.135 = 0.73 atm
P2 = 2X = 0.27 atm
b) fraction of P4 dissociated = X / P4 = 0.135 / 0.865 = 0.156
c)
if volume decreases , the pressure will increase and it shif the reaction towards the reactant according to le chatliers principle