Question

Hydrogen iodide undergoes decomposition according to the equation 2HI(g) H2(g) + I2(g) The equilibrium constant Kp at 500 K for this equilibrium is 0.060. Suppose 0.898 mol of HI is placed in a 1.00-L container at 500 K. What is the equilibrium concentration of H2(g)?

(R = 0.0821 L · atm/(K · mol))

A. 7.3 M

B. 0.40 M

C. 0.15 M

D. 0.18 M

E. 0.043 M

Answer 1

T = 500 K

Δ n = number of gaseous molecule in product - number of gaseous molecule in reactant

Δ n = 0

use:

Kp= Kc (RT)^Δn

0.060 = Kc *(0.0821*500.0)^(0)

Kc = 0.060

initial [HI] = mol of HI / volume

= 0.898 mol / 1.00 L

= 0.898 M

ICE Table:

[HI] [H2] [I2]

initial 0.898 0 0

change -2x +1x +1x

equilibrium 0.898-2x +1x +1x

Equilibrium constant expression is

kc = [H2]*[I2]/[HI]^2

0.06 = (1*x)^2/(0.898-2*x)^2

sqrt(0.06) = (1*x)/(0.898-2*x)

0.2449489742783178 = (1*x)/(0.898-2*x)

0.22-0.4899*x = 1*x

0.22-1.49*x = 0

x = 0.14764

At equilibrium:

[H2] = +1x = +1*0.14764 = 0.14764 M

Answer: 0.15 M