Question

In a 10 Ltr container a mixture of ${H_2}$, ${I_2}$ and $HI$ at equilibrium at ${425^ \circ }C$ there are 0.10 moles of ${H_2}$, 0.10 moles of ${I_2}$ , and 0.74 moles of $HI$. If 0.50 moles $HI$ are now added to this system. What will be the concentration of ${H_2}$, ${I_2}$ and $HI$ once equilibrium has been re-established?

Answer 1

Hint: In a chemical reaction, when two or more reactants are added they react with each other to give a product, and when they reach a state where the concentrations of reactants and products don’t change with time, we say the state of equilibrium is reached.

Complete step by step answer:
As per the question, at equilibrium, it is given that the number of moles of dihydrogen, iodine, and hydrogen iodide is 0.10 moles, 0.10 moles, and 0.74 moles respectively. At the equilibrium the chemical equation will be:
$2HI\overset {} \leftrightarrows {H_2} + {I_2}$
$0.74\;\;\; 0.10 \;\;\; 0.10$
The equilibrium constant ${K_C} = \dfrac{{[{H_2}][{I_2}]}}{{{{[HI]}^2}}}$
$ \Rightarrow {K_C} = \dfrac{{0.10 \times 0.10}}{{{{(0.74)}^2}}}$
$ \Rightarrow {K_C} = 0.0182$
If 0.50 mole of hydrogen iodide is added to the system the total number of moles of hydrogen iodide will be calculated as follows:
Let x moles of hydrogen iodide dissociated into x moles of hydrogen and x moles of iodine at equilibrium then the chemical reaction at equilibrium will be written as:
$2HI\overset {} \leftrightarrows {H_2} + {I_2}$
$ 1.24 \;\;\; 0.10 \;\;\; 0.10$
$(1.24-x)\;\;\; x \;\;\; x $
Now the equilibrium constant will be given as:
${K_C} = \dfrac{{[{H_2}][{I_2}]}}{{{{[HI]}^2}}}$
$ \Rightarrow {K_C} = \dfrac{{x \times x}}{{{{(1.24 - x)}^2}}}$
On putting the value of the equilibrium constant, the above equation becomes:
$ \Rightarrow 0.0182 = \dfrac{{x \times x}}{{{{(1.24 - x)}^2}}}$
$ \Rightarrow 0.0182 = \dfrac{{{x^2}}}{{{{(1.24 - x)}^2}}}$
Or
$ \Rightarrow 0.0182 = {\{ \dfrac{x}{{(1.24 - x)}}\} ^2}$
Taking square root on both side we get:
$ \Rightarrow \sqrt {0.0182} = \dfrac{x}{{(1.24 - x)}}$
$ \Rightarrow 0.1349 = \dfrac{x}{{(1.24 - x)}}$
$ \Rightarrow 0.1349 \times (1.24 - x) = x$
$ \Rightarrow 0.1672 - 0.1349x = x$
$ \Rightarrow 0.1672 = x + 0.1349x$
$ \Rightarrow 0.1672 = 1.1349x$
$ \Rightarrow \dfrac{{0.1672}}{{1.1349}} = x$
$ \Rightarrow x = 0.1473$

Thus, the concentration of hydrogen gas and iodine when equilibrium reestablished is 0.1473 mol/ltr and the concentration of hydrogen iodide is left = 0.124-0.1473= 0.016 mol/ltr

Note: We know that for a reaction equilibrium is reached when the rate of the forward reaction equals the rate of reverse reactions and we have a constant known as equilibrium constant which gives the ratio of products and reactants in a reaction when equilibrium is reached. Also, the equilibrium constant remains the same irrespective of initial concentration.