Question

For a system, H2(g) + I2(g) ⇌ 2 HI(g), Kc = 62.9 at 750 K. 2.80 moles of HI were placed in a 10.0-liter container, brought up to 750 K, and allowed to come to equilibrium. Which situation described below is true, at equilibrium?

a. [HI] = 2 × [H2]
b. [HI] = [H2]
c. [HI] < [H2]
d. [HI] > [H2]
e. [H2] > [I2]

Answer 1

d. [HI] > [H2]

Step-by-step explanation:

The explanation at equilibrium is shown below:-

Data provided
`H_2(g) + I(g) \rightleftharpoons 2HI_(g)`

Initial concentration - -
`(2.80)/(10)` = 0.280 M

At equilibrium x x 0.280 - 2x

`K_c = ((HI)^2)/((H_2)(I_2)) = 62.9`

`= ((0.280 - 2x)^2)/(x^2) = 62.9\\\\4x^2 - 1.12x + 0.0784 = 62.9x^2`

After solve the above equation we will get

x = 0.0282 M

Therefore at equilibrium

`[H_2] = [I_2] = x = 0.0282M\\\\`

`[HI] = 0.280 - 2x = 0.2236 M`

Hence, the correct option is d.