Question

For a given reversible reaction:

H2(g)+I2(g) <====> 2HI(g)

The total pressure of a system at equilibrium is 10 atm. If the number of moles of H2, I2 and HI at equilibrium are 0.5, 0.6 and 0.9 mol respectively and the volume of the vessel is 2 dm3 then verify that Kc=Kp=Kx for the above reversible reaction.

Answer 1

Solution.

$Kx = \frac{n(HI)^2}{n(H2) \times n(I2)}$

$n(X) = \frac{\nu(X)}{\sum(\nu(Xi)}$

n(H2) = 0.25

n(I2) = 0.3

n(HI) = 0.45

Kx = 2.7

$Kp = \frac{p(HI)^2}{p(H2) \times p(I2)}$

$p(X) = n(X) \times p(sys.)$

p(H2) = 2.5 atm

p(I2) = 3 atm

p(HI) = 4.5 atm

Kp = 2.7

$Kc = \frac{C(HI)^2}{C(H2) \times C(I2)}$

$C(X) = \frac{\nu(X)}{V(sys.)}$

C(H2) = 0.25 M

C(I2) = 0.3 M

C(HI) = 0.45 M

Kc = 2.7

Answer:

Kp = Kc = Kx = 2.7