Question

Equal molar amounts of H2 and I2 are placed into an evacuated container and heated to 445°C, where the following equilibrium is established. H2(g) + I2(g) ⇌ 2 HI(g) K = 50.2 Once equilibrium is reached, the concentration of HI is found to be 0.030 M. What were the initial concentrations of H2 and I2 in this experiment?

Answer 1

Answer: The initial concentration of $H_2\text{ and }I_2$ are 0.0192 M and 0.0192 M respectively.

Explanation:

We are given:

Equilibrium concentration of HI = 0.030 M

Moles of hydrogen gas = Moles of iodine gas (concentration will also be the same)

For the given chemical equation:

                  $H_2+I_2\rightleftharpoons 2HI$

Initial:            x     x        -

At eqllm:     x-c    x-c     2c

Calculating the value of 'c'

$2c=0.030\\\\c=\frac{0.030}{2}=0.015M$

The expression of $K_{eq}$ for above reaction follows:

$K_{eq}=\frac{[HI]^2}{[H_2]\times [I_2]}$

We are given:

$K_{eq}=50.2$

$[H_2]=(x-c)=(x-0.015)$

$[I_2]=(x-c)=(x-0.015)$

Putting values in above equation, we get:

$50.2=\frac{(0.030)^2}{(x-0.015)\times (x-0.015)}\\\\x=0.0108,0.0192$

Neglecting the value of x = 0.0108 M, because the initial concentration cannot be less than the equilibrium concentration.

x = 0.0192 M

Hence, the initial concentration of $H_2\text{ and }I_2$ are 0.0192 M and 0.0192 M respectively.