Question

256 g of $HI$ was heated in a sealed bulb at ${{444}^{\circ }}C$ till the equilibrium was attained. The acid was found to be 22 % dissociated at equilibrium. Calculate the equilibrium constant for the reaction.
$2HI(g)\rightleftharpoons {{H}_{2}}(g)+{{I}_{2}}(g)$

Answer 1

Hint: The equilibrium constant of the reaction can be calculated by dividing the concentration of the products in the reaction by the concentration of the reactants in the reaction. Since the dissociation is 22 %, we can convert the dissociation into the number of moles by dividing it by a hundred.

Complete step by step answer:
The molecular mass of hydrogen iodide will be:
$1 + 127 = 128$
Given mass in the question is 256 g, so we can calculate the number of moles will be:
$=\dfrac{256}{128}=2$
Therefore, there are 2 moles of hydrogen iodide in the reaction.
The reaction is:
 $2HI(g)\rightleftharpoons {{H}_{2}}(g)+{{I}_{2}}(g)$
When there is dissociation of 22 % of 2 moles of thee hydrogen iodide then the number of moles will be:
$2\text{ x }\dfrac{22}{100}=0.44$
So, 0.44 moles are dissociated into hydrogen gas and iodine gas. Therefore, the left number of moles of $HI$ will be = 2 – 0.44 = 1.56
In the reaction 2 moles of hydrogen iodide forms 1 mole of hydrogen and 1 mole of iodine gas then, 0.44 mole of hydrogen iodide will form:
$\dfrac{1}{2}\text{ x 0}\text{.44 = 0}\text{.22}$
0.22 mole of hydrogen and 0.22 mole of iodine gas.
- Assuming the reaction takes place in V volume, then the concentration of all the compounds will be:
$[HI]=\dfrac{1.56}{V}$
$[{{H}_{2}}]=\dfrac{0.22}{V}$
$[{{I}_{2}}]=\dfrac{0.22}{V}$
- So, the equilibrium constant of the reaction can be calculated by dividing the concentration of the products in the reaction by the concentration of the reactants in the reaction. This is given below:
${{K}_{c}}=\dfrac{[{{H}_{2}}][{{I}_{2}}]}{[HI]}$
${{K}_{c}}=\dfrac{\dfrac{0.22}{V}\text{ x}\dfrac{0.22}{V}}{\dfrac{1.56}{V}}$
${{K}_{c}}=0.02$
The equilibrium constant is 0.02.

Note: It must be noted that when the equilibrium constant for concentration is calculated then the value of the compounds must be taken in concentration and when the equilibrium constant for pressure is calculated then the value of the compounds must be taken in partial pressure.