Question

For the reaction ${{\text{H}}_{\text{2}}}{\text{ + }}{{\text{I}}_{\text{2}}} \to {\text{2HI}}$ , $K = 47.6$. If the initial number of moles of each reactant and product is one mole, then at equilibrium:
A.${\text{[}}{{\text{I}}_{\text{2}}}{\text{] = [}}{{\text{H}}_{\text{2}}}{\text{], [}}{{\text{I}}_{\text{2}}}{\text{] > [HI]}}$
B.${\text{[}}{{\text{I}}_{\text{2}}}{\text{] = [}}{{\text{H}}_{\text{2}}}{\text{], [}}{{\text{I}}_{\text{2}}}{\text{] < [HI]}}$
C.${\text{[}}{{\text{I}}_{\text{2}}}{\text{] < [}}{{\text{H}}_{\text{2}}}{\text{], [}}{{\text{I}}_{\text{2}}}{\text{] = [HI]}}$
D.${\text{[}}{{\text{I}}_{\text{2}}}{\text{] > [}}{{\text{H}}_{\text{2}}}{\text{], [}}{{\text{I}}_{\text{2}}}{\text{] = [HI]}}$

Answer 1

Hint: Equilibrium constant is defined as the ratio of product of the concentration of products raised to the power of their coefficient in the balanced chemical reaction to the product of the concentration of reactant raised to the power of their coefficient in the balanced chemical reaction.

Complete step by step solution:
First of all let us read about balanced chemical reactions.
Balanced chemical reaction: It is defined as the reaction in which the number of atoms on both the sides i.e. on the reactant side and on the product side should be the same. For example: The reaction given in the question i.e. \[{{\text{H}}_{\text{2}}}{\text{ + }}{{\text{I}}_{\text{2}}} \to {\text{2HI}}\] is balanced chemical reaction because the number of atoms i.e. hydrogen is two on both the sides i.e. on reactant as well as on product side and number of iodine atoms is also two on both the sides i.e. on reactant as well as on product side.
Now if reaction is in forward direction then the number of moles of reactant will decrease and number of moles of product will increase. And similarly when the reaction is in a backward direction then the number of moles of reactant will increase and number of moles of product will decrease.
If we want to calculate the equilibrium constant then it is calculated as:
Equilibrium constant is defined as the ratio of product of the concentration of products raised to the power of their coefficient in the balanced chemical reaction to the product of the concentration of reactant raised to the power of their coefficient in the balanced chemical reaction.
After the reaction let the number of moles which converted to the product is $x$ and in the question we are given with one mole of reactant and product.
So the reaction be
\[
  {{\text{H}}_{\text{2}}}{\text{ + }}{{\text{I}}_{\text{2}}}{\text{ }} \to {\text{ 2HI}} \\
  {\text{1 - x 1 - x 1 + 2x }} \\
 \]
Now the equilibrium constant $K$ will be as $K = \dfrac{{{{(1 + 2x)}^2}}}{{(1 - x)(1 - x)}}$. And the equilibrium constant is given as $K = 47.6$.
Hence,
$
  K = \dfrac{{{{(1 + 2x)}^2}}}{{(1 - x)(1 - x)}} = 47.6 \\
  \dfrac{{1 + 2x}}{{1 - x}} = 6.89 \simeq 7 \\
  1 + 2x = 7 - 7x \\
  9x = 6 \\
  x = \dfrac{2}{3} \\
 $.
And hence the concentration of iodine and hydrogen are same and the concentration of hydrogen iodide is greater than the concentration of iodine i.e. ${\text{[}}{{\text{I}}_{\text{2}}}{\text{] = [}}{{\text{H}}_{\text{2}}}{\text{], [}}{{\text{I}}_{\text{2}}}{\text{] < [HI]}}$.

Hence option B is correct.

Note: For the backward reaction either write the reaction for backward process first then write the formula of equilibrium constant or directly write the equilibrium constant by writing reactants in numerator and products in denominator.