Question

For the equilibrium,
$2{\text{NOCl}}\left( {\text{g}} \right) \rightleftharpoons 2{\text{NO}}\left( {\text{g}} \right) + {\text{C}}{{\text{l}}_2}\left( {\text{g}} \right)$
The value of equilibrium constant ${{\text{K}}_{\text{c}}}$ is ${\text{3}}{\text{.75}} \times {\text{1}}{{\text{0}}^{ - 4}}$ at 1069 K. Calculate the ${{\text{K}}_{\text{p}}}$ for the reaction at this temperature.
A.$0.066$
B.$3.33$
C.$0.03$
D.${\text{0}}{\text{.033}}$

Answer 1

There is a direct relation between ${{\text{K}}_{\text{c}}}$ and ${{\text{K}}_{\text{p}}}$ . We need to put the respective values in the formula. Change in gaseous mole is 1 for the given reaction.
Formula used:
${{\text{K}}_{\text{p}}} = {{\text{K}}_{\text{c}}} \times {\left( {{\text{RT}}} \right)^{\Delta {{\text{n}}_{\text{g}}}}}$
Here ${{\text{K}}_{\text{p}}}$ is equilibrium constant in terms of pressure, ${{\text{K}}_{\text{c}}}$ is equilibrium constant in terms of concentration, R is universal gas constant, T is temperature and $\Delta {{\text{n}}_{\text{g}}}$ is change in gaseous moles or product and reactant.

Complete step by step answer:
Equilibrium constant in terms of concentration that is ${{\text{K}}_{\text{c}}}$ is defined as the ratio of concentration of all products to the concentration of reactant each raised to the power equal to their stoichiometric ratio. Equilibrium constant in terms of pressure that is ${{\text{K}}_{\text{c}}}$ is defined as the ratio of partial pressure of the product raise to the reactant each raised to the power of their stoichiometric coefficient.
For the given reaction:
${{\text{K}}_{\text{c}}} = \dfrac{{{{\left[ {{\text{NO}}} \right]}^2}\left[ {{\text{C}}{{\text{l}}_2}} \right]}}{{{{\left[ {{\text{NOCl}}} \right]}^2}}}$
${{\text{K}}_{\text{p}}} = \dfrac{{{{\left( {{{\text{P}}_{{\text{NO}}}}} \right)}^2} \times {{\text{P}}_{{\text{C}}{{\text{l}}_2}}}}}{{{{\left( {{{\text{P}}_{{\text{NOCl}}}}} \right)}^2}}}$
We have been given the value of ${{\text{K}}_{\text{c}}}$ , temperature and the value of R is $8.314{\text{ J K mo}}{{\text{l}}^{ - 1}}$ .
The change in gaseous mole will be calculated by changing the moles of product minus reactant $2 + 1 - 2 = 1$
Using the formula we will get:
${{\text{K}}_{\text{p}}} = {\text{3}}{\text{.75}} \times {\text{1}}{{\text{0}}^{ - 4}} \times {\left( {{\text{8}}{\text{.314}} \times {\text{1069}}} \right)^1} = 3.33$

Hence, the correct option is B.

Note:
In reversible reactions, at state of equilibrium or when equilibrium is reached the rate of forward reaction becomes equal to rate of backward reaction. That means the rate at which the reactant converts to product becomes equal to the rate at which product converts back to reactant. Also at this stage the number of moles of substance produced per second in the forward direction is equal to the number of moles of substance which disappear per second in the backward direction.