Question

Consider the gas-phase reaction 2 Q (g) ⇌ Z (g) at equilibrium in a

container with a moveable piston at 25 °C. The total pressure in the container is 1.1 atm

and PQ = PZ.


1) Calculate Kp for the reaction at 25 °C

2) The piston is moved to change the volume of the container, resulting in a new total

pressure of 1.4 atm. Assuming no change in temperature, calculate the new partial

pressures of Q and Z when equilibrium is restored

Answer 1

1) $Kp=\frac{Z}{Q^2}$

PQ=PZ

Kp= 1.1/1.1²

Kp= 0.5

2) $Kp=\frac{PZ}{PQ^2}$

Kp=1.4/1.4²

Kp=0.5

but PZ=PQ

P=0.55