Question

Hydrogen iodide decomposes according to the equation 2HI (g)<---> H2 (g) + I2 (g) for which K= .0156 at 400 degrees celsius. If 0.550 mol of HI was injected into 2.00L reaction vessel at 400 degrees celsuis. Calculate the concentration of H2 at equilibrium?

Answer 1

First, find the Keq expression

K = [H2][I2]/[HI]^2

then, we need conditions

initial conditions

[H2] = 0

[I2] = 0

[HI]= mol/V = 0.55/2 = 0.275

in equilibrium

[H2] = 0 + x

[I2] = 0 + x

[HI]= 0.275 - x

now, we can obtain x with K value 0.0156

K = [H2][I2]/[HI]^2

0.0156 = x*x/(0.275-x)

x^2 + 0.0156*x -0.0156*0.275 =0

x = 0.05816

and we know that

[H2] = 0 + x = 0.05816 M

now, we could proof this:

if Q = K, then this is correct

Q = [H2][I2]/[HI]^2

Q = (0.05816*0.05816)/(0.275 - 0.05816) = 0.01559 which is pretty near to 0.0156