Question

At a certain temp, the equilibrium constant for this reaction is 53.3. H2(g)+I2(g)=2HI(g) At this temperature, 0.7 mol of H2 and 0.7 mol of I2 were placed in a 1 L container. What is the concentration of HI present at equilibrium?

Answer 1

Since volume is 1 L value of concentration and number of moles will be same

H2 (g) + I2 (g)    <-------------------> 2HI (g)
0.7               0.7                                             0    (Initial)
0.7-x           0.7-x                                       2x (at equilibrium)

Kc = [HI]^2 / {[H2] [I2]}
53.3 = (2x)^2 / (0.7-x)^2
53.3 = (2x / (0.7-x))^2
2x / (0.7-x) = sqrt (53.3)
2x / (0.7-x) = 7.3
2x = 5.11 - 7.3*x
9.3*x = 5.11
x = 0.55 M

[HI] = 2*x
         = 2*0.55 M
         = 1.1 M
Answer: 1.1 M