Question

HI decomposes to H2 and I2 by the following equation: 2HI(g) → H2(g) + I2(g);Kc = 1.6 × 10−3 at 25∘C If 1.0 M HI is placed into a closed container and the reaction is allowed to reach equilibrium at 25∘C, what is the equilibrium concentration of H2 (g)?

Answer 1

Answer: The concentration of hydrogen gas at equilibrium is 0.037 M

Step-by-step explanation:

We are given:

Initial concentration of HI = 1.0 M

The given chemical equation follows:

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

Initial: 1.0

At eqllm: 1.0-2x x x

The expression of
`K_c` for above equation follows:

`K_c=([H_2][I_2])/([HI]^2)`

We are given:

`Kc=1.6* 10^(-3)`

Putting values in above expression, we get:

`1.6* 10^(-3)=(x* x)/((1.0-2x)^2)\\\\x=-0.043,0.037`

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.037 M

Hence, the concentration of hydrogen gas at equilibrium is 0.037 M