Question

On a given condition, the equilibrium concentration of \[HI,\text{ }{{H}_{2}},\text{ }and\text{ }{{I}_{2}}\]are 0.80, 0.10, and 0.10 mol/litre. The equilibrium constant for the reaction \[{{H}_{2}}+\text{ }{{I}_{2}}\rightleftharpoons 2HI\]will be
(A) 64
(B) 12
(C) 8
(D) 0.8

Answer 1

Hint: Given reaction is at equilibrium position at a particular temperature at this position, rate of forward reaction is equal to rate of backward reaction. In other words, the speed at which reaction is proceeding in forward direction (forming product)is same to which reaction is moving in backward direction (forming reactant). Thus, if we change the concentration of product or reactant, the change will get modified to maintain equilibrium and equilibrium constant and equilibrium constant is the number to which concentration of reactant is multiplied and divided by concentration of product to maintain equilibrium balance.

Formula Used: For general reaction,
\[aA\text{ }+\text{ }bB\rightleftharpoons cC\text{ }+\text{ }dD\], equilibrium constant is defined as
\[Kc=\text{ }{{\left[ C \right]}^{c}}{{\left[ D \right]}^{d}}/{{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}},\text{ }Where\text{ }{{\left[ C \right]}^{c}}and\text{ }{{\left[ D \right]}^{d}}\]are the molar concentration of products, C and D both raised to the power of their stoichiometric coefficients, c and d (number of moles) and similarly for reactant.

Complete Step by Step Solution:
As per hint, to find the value of the equilibrium constant we need the concentration of products and reactants. The chemical equation is \[{{H}_{2}}+\text{ }{{I}_{2}}\rightleftharpoons 2HI\]. Now to find the equilibrium constant for this reaction, the molar concentration of HI (product) and H2 and I2 (reactant) is needed.

The concentration of \[HI,\text{ }{{H}_{2}},\text{ }and\text{ }{{I}_{2}}\]is given, 0.80 mole/litre for\[HI\], 0.10 mol/litre for \[{{H}_{2}}\], and 0.10 mol/litre for \[{{I}_{2}}\]. Now equilibrium constant for a given reaction is
$ {{H}_{2}}+\text{ }{{I}_{2}}\rightleftharpoons 2HI \\$
$ K_c\text{ }=\text{ }\left[ HI \right]{}^\text{2}/[{{H}_{2}}][{{I}_{2}}] \\$
$ K_c\text{ }=\text{ }0.80\text{ }\times \text{ }0.80/0.10\text{ }\times \text{ }0.10 \\$
$ K_c\text{ }=\text{ }64 \\$
Thus, the correct option is A.

Note: In this question molar concentration of both reactant and products is given but in some questions, we only get moles of reactant and product and also volume in litre. In that case, we need to find the molar concentration of both reactants and products. Also, stoichiometry coefficients are different from the actual coefficient. The stoichiometry coefficient is the moles used to balance the chemical equation.