Question

An equilibrium system for the reaction b/w Hydrogen and Iodine to give hydrogen iodide at $670K$ in a 5-liter flask contains $0.4$moles of hydrogen, $0.4$moles of Iodine, and $2.4$ mole of Hydrogen Iodide. Calculate the equilibrium constant.

Answer 1

Equilibrium constant is the rate at which a substance reacts is proportional to its molar concentration and the rate of the chemical reaction is proportional to the product of the molar concentration of the reacting substances.

Formula used: ${K_c} = \dfrac{{\left[ {product} \right]}}{{\left[ {reac\tan t} \right]}}$
Here, ${K_c}$ = equilibrium constant
$\left[ {product} \right]$ = concentration of product
$\left[ {reac\tan t} \right]$ = concentration of reactant

Complete step by step answer:
Hydrogen and iodide react to form hydrogen iodide.
Let us write the balanced chemical reaction for hydrogen iodide:
${H_2} + {I_2} \rightleftarrows 2HI$
At equilibrium, we are given $0.4$moles of hydrogen, $0.4$moles of Iodine, and $2.4$ mole of Hydrogen Iodide. And the volume of the flask i.e. 5 liters.
Now, to find the equilibrium constant we have to convert the given number of moles into molar concentration.
Concentration is the number of moles divided by the volume.
$\left[ {{\text{Concentration}}} \right]{\text{ = }}\dfrac{{{\text{number of moles}}}}{{{\text{volume in liter}}}}$
Now let us find the concentration of hydrogen at equilibrium:
$\left[ {{H_2}} \right] = \dfrac{{0.4mole}}{{5L}} = 0.08mol{L^{ - 1}}$
Let us find the concentration of iodine at equilibrium:
$\left[ {{I_2}} \right] = \dfrac{{0.4mole}}{{5L}} = 0.08mol{L^{ - 1}}$
Let us find the concentration of hydrogen iodide at equilibrium:
$\left[ {HI} \right] = \dfrac{{2.4mole}}{{5L}} = 0.48mol{L^{ - 1}}$
Now the equilibrium constant for the hydrogen iodide reaction is given as follows:
${K_c} = \dfrac{{{{\left[ {HI} \right]}^2}}}{{\left[ {{H_2}} \right]\left[ {{I_2}} \right]}}$
Let us substitute for each concentration:
${K_c} = \dfrac{{{{\left[ {0.48} \right]}^2}}}{{\left[ {0.08} \right]\left[ {0.08} \right]}}$
By opening the bracket we get:
${K_c} = \dfrac{{0.2304}}{{0.0064}}$
By dividing we get the value of the equilibrium constant:
${K_c} = 36$

Therefore, we can conclude that the equilibrium constant for the following reaction is 36.

Note: In this type of question we must always try to find the concentration of reactant and product at the equilibrium. We must also write the balanced chemical equation because equilibrium concentration also depends on the stoichiometric of the reactant and product.