Question

A closed system initially containing 1.000 × 10-3 M H2 and 2.000 × 10-3 M I2 at 448℃ is allowed to reach equilibrium. At equilibrium, the HI concentration is 1.87 × 10-3 M. Given that the reaction equation is H2(g) + I2(g) ↔ 2 HI(g), find:
a. Equilibrium concentrations of the two reactants
b. Kc of the reaction taking place.

Answer 1

Solution:

Chemical reaction:

H₂(g) + I₂(g) ↔ 2HI(g)

Given:

[H]_in = 10^-3M

[I]_in = 2*10^-3M

[HI]_eq = 1.87*10^-3M

this problem is solved by using ICE table method:

$\def\arraystretch{1.5} \begin{array}{c:c:c} H_2 & I_2 & 2HI \\ \hline 1*10^{-3}M & 2*10^{-3}M & 0 \\ \hdashline x & x & 2x\\ \hdashline [H_{2}]_{eq} & [I_{2}]_{eq} & 1.87*10^{-3}M \end{array}$

Firstly we should find x(change) in the ICE table:

2x=1.87*10^-3

x=9.35*10^-4

a) Equilibrium concentrations of the two reactants:

[H]_eq = 10^-3 - 9.35*10^-4 = 6.5 * 10^-5 M

[I]_eq = 2*10^-3 - 9.35*10^-4 = 1.065 * 10^-3 M

b) Kc of the reaction taking place:

$K_c= \dfrac{[HI]^2}{[H][I]} = \dfrac{(1.87*10^{-3})^2}{(1.065*10^{-3})(6.5*10^{-5})} =50.5$

$K_c=50.5$