Question

The reaction X 2 (g) m 2 X(g) occurs in a closed reaction vessel at constant volume and temperature. Initially, the vessel contains only X 2 at a pressure of 1.55 atm. After the reaction reaches equilibrium, the total pressure is 2.85 atm. What is the value of the equilibrium constant, Kp , for the reaction?

Answer 1

Answer:

41.6 is the value of the equilibrium constant for the reaction.

Explanation:

           $X_2\rightleftharpoons 2X$

Initially   1.55            0

At eq'm  1.55-p        2p

Total pressure at the equilibrium = P =2.85 atm

$P=(1.55-p)+2 p = 2.85 atm$

p = 1.3 atm

Partial pressure of $X_2$ at equilibrium:

$[p_{X_2}^o]=2p=2\time 1.3 atm=2.6 atm$

Partial pressure of X at equilibrium;

$[p_{X}^{o}]=1.55 atm -1.3 atm = 0.25 atm$

The value of equilibrium constant will be given as:

$K_p=\frac{[p_{X_2}^o]}{[p_{X}^{o}]^2}$

$K_p=\frac{2.6 atm}{(0.25 atm)^2}=41.6$