Answer:
41.6 is the value of the equilibrium constant for the reaction.
Explanation:
[tex]X_2\rightleftharpoons 2X[/tex]
Initially 1.55 0
At eq'm 1.55-p 2p
Total pressure at the equilibrium = P =2.85 atm
[tex]P=(1.55-p)+2 p = 2.85 atm[/tex]
p = 1.3 atm
Partial pressure of [tex]X_2[/tex] at equilibrium:
[tex][p_{X_2}^o]=2p=2\time 1.3 atm=2.6 atm[/tex]
Partial pressure of X at equilibrium;
[tex][p_{X}^{o}]=1.55 atm -1.3 atm = 0.25 atm[/tex]
The value of equilibrium constant will be given as:
[tex]K_p=\frac{[p_{X_2}^o]}{[p_{X}^{o}]^2}[/tex]
[tex]K_p=\frac{2.6 atm}{(0.25 atm)^2}=41.6[/tex]