Question

Hydrogen gas, iodine vapor, hydrogen iodine are mixed in a flask and heated to 617°C. H2(g) + I2(g) ⇋ 2 HI(g) Kc = 47 at 617°C If the initial concentrations of hydrogen gas and iodine vapor are both 0.041 mol/L and the concentration of hydrogen iodine is 0.115 mol/L what is the equilibrium concentration of hydrogen gas? Enter a number to 4 decimal places.

Answer 1

Answer: The equilibrium concentration of hydrogen gas is 0.0222 M

Explanation:

We are given:

Initial concentration of hydrogen gas = 0.041 M

Initial concentration of iodine vapor = 0.041 M

Initial concentration of hydrogen iodide = 0.115 M

For the given chemical equation:

                     $H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$

Initial:            0.041     0.041     0.115

At eqllm:     0.041-x   0.041-x   0.115+2x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

We are given:

$K_c=47$

Putting values in above equation, we get:

$47=\frac{(0.115+2x)^2}{(0.041-x)\times (0.041-x)}\\\\x=0.0816,0.0188$

Neglecting the value of x = 0.0816 because, equilibrium concentration cannot be greater than the initial concentration

So, equilibrium concentration of hydrogen gas = $(0.041-x)=(0.041-0.0188)=0.0222M$

Hence, the equilibrium concentration of hydrogen gas is 0.0222 M