Question

The value of $ {K_p}{\text{ = 2}}{\text{.0 }} \times {\text{ 1}}{{\text{0}}^{10}}{\text{ bar}} $ is given at $ 450{\text{ K}} $ for the given reaction at equilibrium.
 $ {\text{2 S}}{{\text{O}}_2}(g){\text{ + }}{{\text{O}}_2}(g){\text{ }} \rightleftharpoons {\text{ 2 S}}{{\text{O}}_3}(g) $
What is the value of $ {K_c} $ at the same temperature?
 $ (i){\text{ 7}}{\text{.4 }} \times {\text{ 1}}{{\text{0}}^{11}}{\text{ L mo}}{{\text{l}}^{ - 1}} $
 $ (ii){\text{ 7}}{\text{.4 }} \times {\text{ 1}}{{\text{0}}^{ - 11}}{\text{ L mo}}{{\text{l}}^{ - 1}} $
 $ (iii){\text{ 3}}{\text{.7 }} \times {\text{ 1}}{{\text{0}}^{ - 11}}{\text{ L mo}}{{\text{l}}^{ - 1}} $
 $ (iv){\text{ 3}}{\text{.7 }} \times {\text{ 1}}{{\text{0}}^{11}}{\text{ L mo}}{{\text{l}}^{ - 1}} $

Answer 1

Hint :We will find the difference between the number of gaseous products and reactants. With the help of this change in gaseous number of moles we will find the value of $ {K_c} $ from the relation between $ {K_p} $ and $ {K_c} $ for a reaction at equilibrium.
For finding $ {K_p} $ or $ {K_c} $ :
 $ {K_p}{\text{ = }}{K_c}{\text{ }} \times {\text{ }}{\left( {{\text{RT}}} \right)^{\Delta {n_g}}} $
Here, $ R $ is gas constant, $ T $ is the temperature at which the reaction takes place and $ \Delta {n_g} $ is the change in the number of gaseous moles of the reaction.

Complete Step By Step Answer:
For a reaction at equilibrium we can find the value of $ {K_p} $ or $ {K_c} $ by using the relation between them as,
 $ {K_p}{\text{ = }}{K_c}{\text{ }} \times {\text{ }}{\left( {{\text{RT}}} \right)^{\Delta {n_g}}} $
Firstly we will find the change in number of gaseous moles as:
 $ \Delta {n_g}{\text{ = }}{n_{g({\text{products}})}}{\text{ - }}{n_{g({\text{reactants}})}} $
For the given reaction it can be calculated as,
 $ {\text{2 S}}{{\text{O}}_2}(g){\text{ + }}{{\text{O}}_2}(g){\text{ }} \rightleftharpoons {\text{ 2 S}}{{\text{O}}_3}(g) $
 $ \Delta {n_g}{\text{ = }}{n_{g({\text{products}})}}{\text{ - }}{n_{g({\text{reactants}})}} $
We can observe that the number of gaseous moles of the product is two while the total number of gaseous moles of reactant is three. Therefore substituting these value we get change in number of gaseous moles for the whole reaction as,
 $ \Delta {n_g}{\text{ = 2 - 3 = - 1}} $
 Hence $ \Delta {n_g}{\text{ = - 1}} $ . Now substituting the all given values we get the result as,
 $ {K_p}{\text{ = }}{K_c}{\text{ }} \times {\text{ }}{\left( {{\text{RT}}} \right)^{\Delta {n_g}}} $
On putting, $ {K_p}{\text{ = 2}}{\text{.0 }} \times {\text{ 1}}{{\text{0}}^{10}}{\text{ bar}} $ , $ R{\text{ = 0}}{\text{.0831 L bar }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}} $ , $ T = {\text{ 450 K}} $ and $ \Delta {n_g}{\text{ = - 1}} $
 $ {\text{2}}{\text{.0 }} \times {\text{ 1}}{{\text{0}}^{10}}{\text{ bar = }}{K_c}{\text{ }} \times {\text{ }}{\left( {{\text{0}}{\text{.0831 L bar }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}{\text{ }} \times {\text{ 450 K}}} \right)^{ - 1}} $
 $ {{\text{K}}_c}{\text{ = }}\left( {{\text{2 }} \times {\text{ 1}}{{\text{0}}^{10}}{\text{ bar}}} \right){\text{ }} \times \left( {{\text{0}}{\text{.0831 L bar }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}{\text{ }} \times {\text{ 450 K}}} \right) $
 $ {{\text{K}}_c}{\text{ = 7}}{\text{.4 }} \times {\text{ 1}}{{\text{0}}^{11}}{\text{ L mo}}{{\text{l}}^{ - 1}} $
Therefore the value of $ {K_c} $ for the given reaction at equilibrium is equal to $ {\text{7}}{\text{.4 }} \times {\text{ 1}}{{\text{0}}^{11}}{\text{ L mo}}{{\text{l}}^{ - 1}} $

Therefore the correct option is $ (i){\text{ 7}}{\text{.4 }} \times {\text{ 1}}{{\text{0}}^{11}}{\text{ L mo}}{{\text{l}}^{ - 1}} $ .

Note :
We cannot use the above relation between $ {K_p} $ and $ {K_c} $ when the temperature of reaction is not constant. If the temperature varies then we cannot use this relation. The value of gas constant $ R $ must be chosen correctly according to given units of $ {K_p} $ . If we use other values of gas constant then our answer would be different. We must include gaseous moles only while finding $ \Delta {n_g} $ for the reaction.