Question

Calculate ${{\text{K}}_{\text{c}}}$ and ${{\text{K}}_{\text{P}}}$ at $500{\text{K}}$ for the reaction, $2HI(g) \rightleftharpoons {H_2}(g) + {I_2}(g)$ if the equilibrium concentrations are $\left[ {{\text{HI}}} \right] = 0.5{\text{M}}$, $\left[ {{{\text{H}}_{\text{2}}}} \right] = 0.08{\text{M}}$ and $\left[ {{{\text{I}}_{\text{2}}}} \right] = 0.062{\text{M}}$.

Answer 1

To answer this question, you must recall the formula of equilibrium constant of a reaction in terms of the concentration of reactants, that is ${{\text{K}}_{\text{c}}}$. It is given by the concentration at equilibrium of products divided by the concentration at equilibrium of reactants raised to the power of their respective stoichiometric coefficients.
Formula used:
${{\text{K}}_{\text{c}}} = \dfrac{{\left[ {{{\text{H}}_{\text{2}}}} \right]\left[ {{{\text{I}}_{\text{2}}}} \right]}}{{{{\left[ {{\text{HI}}} \right]}^{\text{2}}}}}$
Where ${{\text{K}}_{\text{c}}}$is equilibrium constant of the reaction
$\left[ {{\text{HI}}} \right]$is the concentration of ${\text{HI}}$ at equilibrium
$\left[ {{{\text{H}}_2}} \right]$is the concentration of ${{\text{H}}_2}$ at equilibrium
$\left[ {{{\text{I}}_2}} \right]$ is the concentration of ${{\text{I}}_2}$ at equilibrium

Complete step by step answer:
For the given reaction, $2HI(g) \rightleftharpoons {H_2}(g) + {I_2}(g)$.
We are given the equilibrium concentrations of the reactants as well as the products.
We have, $\left[ {{\text{HI}}} \right] = 0.5{\text{M}}$, $\left[ {{{\text{H}}_{\text{2}}}} \right] = 0.08{\text{M}}$ and $\left[ {{{\text{I}}_{\text{2}}}} \right] = 0.062{\text{M}}$at equilibrium.
To find ${{\text{K}}_{\text{c}}}$, we use the formula ${{\text{K}}_{\text{c}}} = \dfrac{{\left[ {{{\text{H}}_{\text{2}}}} \right]\left[ {{{\text{I}}_{\text{2}}}} \right]}}{{{{\left[ {{\text{HI}}} \right]}^{\text{2}}}}}$.
Substituting the values of concentrations into the equation, we get,
${{\text{K}}_{\text{c}}} = \dfrac{{\left( {0.08} \right) \times \left( {0.062} \right)}}{{{{\left( {0.5} \right)}^2}}}$
$\Rightarrow {{\text{K}}_{\text{c}}} = \dfrac{{0.00496}}{{0.25}}$
Solving this, we get:
$\therefore {{\text{K}}_{\text{c}}} = 0.0198$
Therefore, the equilibrium constant of the reaction in terms of the concentrations of the constituents is ${{\text{K}}_{\text{c}}} = 0.0198$.
Now we have to find the equilibrium constant of the reaction in terms of the partial pressures of the gaseous constituents, that is ${{\text{K}}_{\text{P}}}$.
We know the relation between ${{\text{K}}_{\text{c}}}$ and ${{\text{K}}_{\text{P}}}$ is
${{\text{K}}_{\text{P}}} = {{\text{K}}_{\text{c}}} \times {\left( {{\text{RT}}} \right)^{\Delta {n_g}}}$
Where, ${\text{R}}$is the gas constant
${\text{T}}$is the temperature
We are given the temperature $= 500{\text{K}}$
$\Delta {{\text{n}}_{\text{g}}}$is the change in number of moles of gaseous constituents,
Which is the difference between the number of moles of gases in the product and the number of moles of gases in the reactant.
Thus, $\Delta {{\text{n}}_{\text{g}}} = 2 - 2 = 0$
Substituting the values in the relation, we get
${{\text{K}}_{\text{P}}} = {{\text{K}}_{\text{c}}} \times {\left( {500{\text{R}}} \right)^0}$
Simplifying:
$\Rightarrow {{\text{K}}_{\text{P}}} = {{\text{K}}_{\text{c}}}$
$\therefore {{\text{K}}_{\text{P}}} = 0.0198$

Note:
The numerical value of an equilibrium constant is obtained by allowing a reaction to proceed to equilibrium and then measuring the concentrations of each substance involved in that reaction. Since the concentrations are measured at equilibrium, the equilibrium constant remains the same for a given reaction independent of initial concentrations of the reactants and products. This knowledge allows us to derive a model expression that can serve as a standard for any reaction. This basic standard form of an equilibrium constant is given as
${K_{eq}} = \dfrac{{{{\left( {{a_C}} \right)}^c}}}{{{{\left( {{a_A}} \right)}^a}{{\left( {{a_B}} \right)}^b}}}$ for a reaction: $aA + bB \rightleftharpoons cC$
If $K > 1$ then equilibrium favours the formation of products, reaction proceeds in the forward direction.
If $K < 1$ then equilibrium favours the formation of reactants, reaction proceeds in the backward direction.