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

Answer:

d. [HI] > [H2]

Explanation:

The explanation at equilibrium is shown below:-

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

Initial concentration    -           -           $\frac{2.80}{10}$ = 0.280 M

At equilibrium             x          x       0.280 - 2x

$K_c = \frac{(HI)^2}{(H_2)(I_2)} = 62.9$

$= \frac{(0.280 - 2x)^2}{x^2} = 62.9\\\\4x^2 - 1.12x + 0.0784 = 62.9x^2$

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.