Question

In the reaction between the hydrogen and iodine, 6.34 moles of hydrogen and 4.02 moles of iodine are found to be present equilibrium with 42.85 moles of hydrogen iodine at ${35^ \circ }c$. Calculate the equilibrium constant?
(A) 72.04
(B) 65.38
(C) 52.08
(D) 45.20

Answer 1

We should know the chemical equation of the reaction between hydrogen and iodine. The relationship between the amount of reactant and product when it is at equilibrium is equal to the equilibrium constant.

Complete step by step answer:
The reaction between the hydrogen and iodine is given by:
${H_2} + {I_2}\underset {} \leftrightarrows 2HI$
Number of moles in equilibrium: 6.34 4.02 42.85

We need to calculate molar concentration, $$Molar{\text{ }}concentration = \dfrac{{No.of.moles}}{{volume}}$$ Molar concentration is given by the ratio of moles of a solute and liters in volume Let the volume of ${H_2}$, ${I_2}$ and HI be V, $$\;Molar{\text{ }}concentration{\text{ }}of\;HI = \dfrac{{42.85}}{V}$$ $$\;Molar{\text{ }}concentration{\text{ }}of\;{I_2} = \dfrac{{4.02}}{V}$$ $$\;Molar{\text{ }}concentration{\text{ }}of\;{H_2} = \dfrac{{6.34}}{V}$$ The formula to calculate equilibrium constant: $${k_c} = \dfrac{{[product]}}{{[reactant]}}$$ Equilibrium constant is the ratio between molar concentration of product and molar concentration of reactant. $${k_c} = \dfrac{{{{[HI]}^2}}}{{[{H_2}][{I_2}]}}$$ $${k_c} = \dfrac{{{{[\dfrac{{42.85}}{V}]}^2}}}{{[\dfrac{{6.34}}{V}][\dfrac{{4.02}}{V}]}}$$ $${k_c} = \dfrac{{{{[42.85]}^2}}}{{25.4868}}$$ $${k_c} = 72.04$$ **Thus, the correct option is A** **Note:** Firstly we should be thorough with formula, $${k_c} = \dfrac{{[product]}}{{[reactant]}}$$ Here, $${k_c} = \dfrac{{{{[HI]}^2}}}{{[{H_2}][{I_2}]}}$$ We should make sure that we don't forget the power of the product in this case while calculating. If we forget squares, that will result in wrong answers.