Question

2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to reach equilibrium. Calculate the concentration of H2 at equilibrium. 0.0275 M (Your correct answer) 0.138 M 0.275 M 0.550 M 0.220 M

Answer 1

Answer: The concentration of hydrogen gas at equilibrium is 0.0275 M

Step-by-step explanation:

Molarity is calculated by using the equation:

`\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}`

Moles of HI = 0.550 moles

Volume of container = 2.00 L

`\text{Initial concentration of HI}=(0.550)/(2)=0.275M`

For the given chemical equation:

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

Initial: 0.275

At eqllm: 0.275-2x x x

The expression of
`K_c` for above equation follows:

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

We are given:

`K_c=0.0156`

Putting values in above expression, we get:

`0.0156=(x* x)/((0.275-2x)^2)\\\\x=-0.0458,0.0275`

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

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M