Question

For the reaction ${H_2} + {I_2} \to 2HI$ , the ${K_c}$ is $49$ . Calculate the concentration of $HI$ at equilibrium when initially one mole of ${H_2}$ is mixed with one mole of ${I_2}$ in $2$ litre flask?

Answer 1

The equilibrium constant from the concentrations was represented by ${K_c}$ , which is the ratio of the concentrations of product to concentration of reactants. Equating the given value of ${K_c}$ with the concentrations of products and reactants at equilibrium gives the change of concentration. The concentration can be determined from the ratio of moles to given volume.

Complete answer:
Given that for the reaction ${H_2} + {I_2} \to 2HI$ , the ${K_c}$ is $49$
Given that initially one mole of ${H_2}$ is mixed with one mole of ${I_2}$ in $2$ litre flask.
Thus, the concentration will be $\dfrac{{1mol}}{{2L}} = 0.5M$
Based on the below ICE table,

| ${H_2}$ | ${I_2}$ | $HI$
---|---|---|---
initial| $0.5$ | $0.5$ | $0$
change| $- x$ | $- x$ | $+ 2x$
Equilibrium| $0.5 - x$ | $0.5 - x$ | $+ 2x$

By equating the ratio of concentrations of products to ratio with the given value of ${K_c}$ , which is given as $49$
$\dfrac{{{{\left( {2x} \right)}^2}}}{{\left( {0.5 - x} \right)\left( {0.5 - x} \right)}} = 49$
By solving the above equation,
$4{x^2} = 49\left( {0.25 - x + {x^2}} \right)$
The quadratic equation is written as $45{x^2} - 49x + 12.25 = 0$
To find out the roots of the above quadratic equation, it can be written as
$x = \dfrac{{49 \pm \sqrt {{{\left( { - 49} \right)}^2} - 4\left( {45} \right)\left( {12.25} \right)} }}{{2\left( {45} \right)}}$
The value of $x$ will be equal to $0.7$ or $0.389$
As the concentration at equilibrium is less than $0.5$ , the value $0.7$ can be neglected.
Thus, the value of $x$ will be $0.389$
The concentration of $HI$ at equilibrium will be $2\left( {0.389} \right) = 0.778M$
Thus, the concentration of $HI$ at equilibrium when initially one mole of ${H_2}$ is mixed with one mole of ${I_2}$ in $2$ litre flask is $0.778M$ .

Note:
Molarity is also known as molar concentration which is the ratio of the moles to the volume of solution in litres. When the volume is in millilitres, change it into litres. Based on obtained molar concentration only, the concentrations at equilibrium can be determined.