Question

At 700K, the equilibrium constant, ${{\text{K}}_{\text{p}}}$ for the reaction ${\text{2S}}{{\text{O}}_{\text{3}}}\left( {\text{g}} \right)\underset {} \leftrightarrows {\text{2S}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right){\text{ + }}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$ is ${\text{1}}{{.80 \times 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{kpa}}$ . What is the numerical value in moles per litre of ${{\text{K}}_{\text{c}}}$ for this reaction at the same temperature?
A.$3.09\times 10^{-7}molL^{-1}$
B.$3.09\times 10^{-6}molL^{-1}$
C.$3.09\times 10^{7}molL^{-1}$
D.$3.09\times 10^{6}molL^{-1}$

Answer 1

Hint: According to the law of chemical equilibrium, the product of the molar concentrations of the product species, each raised to the power equal to its coefficient divided by the product of the molar concentrations of the reactant species, each raised to the power equal to its coefficient at constant temperature is constant. This constant is called the equilibrium constant.
If we consider a general reaction at the state of equilibrium
${\text{aA + bB}}\overset {} \leftrightarrows {\text{cC + dD}}$
The value of the equilibrium constant, ${{\text{K}}_{\text{c}}}$ can be written as:
${{\text{K}}_{\text{c}}} = \left( {\dfrac{{{{\left[ {\text{C}} \right]}^{\text{c}}}{{\left[ {\text{D}} \right]}^{\text{d}}}}}{{{{\left[ {\text{A}} \right]}^{\text{a}}}{{\left[ {\text{B}} \right]}^{\text{b}}}}}} \right)$
This expression represents the law of chemical equilibrium.
For gas phase reactions, the equilibrium constant can also be expressed in terms of partial pressures of the reactants and products and in that case, the equilibrium constant is denoted by ${{\text{K}}_{\text{p}}}$ .

Complete step by step answer:
Given that, at 700K, the equilibrium constant, ${{\text{K}}_{\text{p}}}$ for the ${\text{2S}}{{\text{O}}_{\text{3}}}\left( {\text{g}} \right)\underset {} \leftrightarrows {\text{2S}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right){\text{ + }}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$ reaction is ${\text{1}}{{.80 \times 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{kpa}}$ .
We need to find out the value of ${{\text{K}}_{\text{c}}}$ for this reaction at the same temperature.
For some reactions the values of ${{\text{K}}_{\text{p}}}$ and ${{\text{K}}_{\text{c}}}$ are equal. But for many other reactions, they have different values and hence it is preferable to calculate one from the other.
The relation between ${{\text{K}}_{\text{c}}}$and ${{\text{K}}_{\text{p}}}$ is given by the following expression:
\[{{\text{K}}_{\text{c}}}{\text{ = }}{{\text{K}}_{\text{p}}} \times \dfrac{{\text{1}}}{{{{\left( {{\text{RT}}} \right)}^{\Delta {\text{n}}}}}}\]
Here, $\Delta {\text{n}}$ represents the difference between the number of moles of gaseous products and that of gaseous reactants. The numerical value of $\Delta {\text{n}}$ is obtained from the coefficients of the balanced equation.
R and T is the gas constant and temperature respectively.
The value of R is $0.0821{\text{Latm}}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ . So, the value of ${{\text{K}}_{\text{p}}}$ is also converted to atm.
Since 1pa is equal to ${10^{ - 5}}{\text{atm}}$ , therefore
$
  {{\text{K}}_{\text{p}}} = 1.8 \times {10^{ - 3}}{\text{kpa}} \\
   \Rightarrow {{\text{K}}_{\text{p}}} = 1.8{\text{kpa}} \\
   \Rightarrow {{\text{K}}_{\text{p}}} = 1.8 \times {10^{ - 5}}{\text{atm}} \\
 $
For the given reaction, $\Delta {\text{n = }}\left( {2 + 1} \right) - 2 = 1$
Substitute all these in the expression for relation between ${{\text{K}}_{\text{c}}}$ and ${{\text{K}}_{\text{p}}}$.
\[
  {{\text{K}}_{\text{c}}}{\text{ = 1}}{\text{.8}} \times {\text{1}}{{\text{0}}^{ - 5}} \times \dfrac{{\text{1}}}{{{{\left( {0.0821 \times {\text{700}}} \right)}^1}}}{\text{mol}}{{\text{L}}^{{\text{ - 1}}}} \\
   \Rightarrow {{\text{K}}_{\text{c}}}{\text{ = }}0.0309 \times {\text{1}}{{\text{0}}^{ - 5}}{\text{mol}}{{\text{L}}^{{\text{ - 1}}}} \\
   \Rightarrow {{\text{K}}_{\text{c}}}{\text{ = }}3.09 \times {10^{ - 7}}{\text{mol}}{{\text{L}}^{{\text{ - 1}}}} \\
 \]

Thus, the correct option is A.

Note:
-The value of the equilibrium constant is independent of the original concentrations of the reactants.
-The value of the equilibrium constant is independent of the presence of a catalyst.
-Equilibrium constant of the forward reaction is the inverse of the equilibrium constant of the reverse reaction.