Question

Be sure to answer all parts. Hydrogen iodide decomposes according to the reaction 2 HI(g) ⇌ H2(g) + I2(g) A sealed 1.50−L container initially holds 0.00623 mol of H2, 0.00414 mol of I2, and 0.0244 mol of HI at 703 K. When equilibrium is reached, the concentration of H2(g) is 0.00467 M. What are the concentrations of HI(g) and I2(g)?

Answer 1

Answer : The concentration of $HI$ and $I_2$ at equilibrium is, 0.0158 M and 0.00302 M respectively.

Explanation :

First we have to calculate the concentration of $H_2, I_2\text{ and }HI$

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.00623mol}{1.50L}=0.00415M$

$\text{Concentration of }I_2=\frac{\text{Moles of }I_2}{\text{Volume of solution}}=\frac{0.00414mol}{1.50L}=0.00276M$

$\text{Concentration of }HI=\frac{\text{Moles of }HI}{\text{Volume of solution}}=\frac{0.0244mol}{1.50L}=0.0163M$

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

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

Initial conc.      0.0163     0.00415      0.00276

At eqm.        (0.0163-2x) (0.00415+x)  (0.00276+x)

As we are given:

Concentration of $H_2$ at equilibrium = 0.00467 M

That means,

(0.00415+x) = 0.00467

x = 0.00026 M

Concentration of $HI$ at equilibrium = (0.0163-2x) = (0.0163-2(0.00026)) = 0.0158 M

Concentration of $I_2$ at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M