Question

Units of ${{\text{K}}_{\text{c}}}$ for ${\text{xA(g)}} \rightleftharpoons {\text{yB(g)}}$ is ${\text{lit}}{\text{.mo}}{{\text{l}}^{{\text{ - 1}}}}$ then the value of ${\text{x}}$ and ${\text{y}}$ are;
A. ${\text{1,2}}$
B. ${\text{3,2}}$
C. ${\text{2,3}}$
D. None

Answer 1

${{\text{K}}_{\text{c}}}$ is an equilibrium constant in terms of molar concentrations of products and reactants which are involved in the chemical reaction. If the reactants are present in the gaseous state, the concentrations are replaced by partial pressure of the substances.

Complete answer:
As we know that the unit of ${{\text{K}}_{\text{c}}}$ is ${\text{mol}}{\text{.li}}{{\text{t}}^{{\text{ - 1}}}}$.
For the given reaction,
${\text{xA(g)}} \rightleftharpoons {\text{yB(g)}}$
Let’s write the ${{\text{K}}_{\text{c}}}$expression for the given reaction.
${{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{{{\text{K}}_{\text{f}}}}}{{{{\text{K}}_{\text{b}}}}}$
${{\text{K}}_{\text{f}}}$ = Equilibrium expression for the forward reaction.
${{\text{K}}_{\text{b}}}$ = Equilibrium expression for the backward reaction.
Let’s write the ${{\text{K}}_{\text{c}}}$expression for the given reaction.
${{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{{{{\text{[B]}}}^{\text{y}}}}}{{{{{\text{[A]}}}^{\text{x}}}}}$
Now using the hit and trial method we will try to find the values of ${\text{x}}$and ${\text{y}}$for which the given units are satisfied.

For option - ${\text{A}}$-${\text{x = 1 and y = 2}}$
${{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{{{{\text{[B]}}}^{\text{y}}}}}{{{{{\text{[A]}}}^{\text{x}}}}}$
Substitute the ${\text{x and y}}$values.
${{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{{{{\text{(mol}}{\text{.}}{{\text{l}}^{{\text{ - 1}}}}{\text{)}}}^{\text{2}}}}}{{{{{\text{(mol}}{\text{.}}{{\text{l}}^{{\text{ - 1}}}}{\text{)}}}^{\text{1}}}}}{\text{ = (mol}}{\text{.}}{{\text{l}}^{{\text{ - 1}}}}{\text{)}}$

For option - ${\text{B}}$- ${\text{x = 3 and y = 2}}$
${{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{{{{\text{[B]}}}^{\text{y}}}}}{{{{{\text{[A]}}}^{\text{x}}}}}$
Substitute the ${\text{x and y}}$values.

$${{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{{{{\text{[B]}}}^{\text{2}}}}}{{{{{\text{[A]}}}^{\text{3}}}}} \\\ \Rightarrow {{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{{{{\text{(mol}}\,.{{\text{l}}^{{\text{ - 1}}}}{\text{)}}}^{\text{2}}}}}{{{{{\text{(mol}}\,.{{\text{l}}^{{\text{ - 1}}}}{\text{)}}}^{\text{3}}}}} \\\ \Rightarrow {{\text{K}}_{\text{c}}}{\text{ = (mol}}\,.{{\text{l}}^{{\text{ - 1}}}}{{\text{)}}^{{\text{ - 1}}}} \\\ \Rightarrow {{\text{K}}_{\text{c}}}{\text{ = (lit}}{\text{.}}\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}{\text{)}} \\\$$

For option - ${\text{C}}$- ${\text{x = 2 and y = 3}}$
${{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{{{{\text{[B]}}}^{\text{y}}}}}{{{{{\text{[A]}}}^{\text{x}}}}}$
Substitute the ${\text{x and y}}$ values.

$${{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{{{{\text{[B]}}}^{\text{3}}}}}{{{{{\text{[A]}}}^{\text{2}}}}} \\\ \Rightarrow {{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{{{{\text{(mol}}\,{{\text{l}}^{{\text{ - 1}}}}{\text{)}}}^{\text{3}}}}}{{{{{\text{(mol}}\,{{\text{l}}^{{\text{ - 1}}}}{\text{)}}}^{\text{2}}}}} \\\ \Rightarrow {{\text{K}}_{\text{c}}}{\text{ = (mol}}\,{{\text{l}}^{{\text{ - 1}}}}{\text{)}} \\\$$

From the above solution unit of ${{\text{K}}_{\text{c}}}$ obtained from option ${\text{B}}$.
Hence, the correct option is ${\text{B}}$.

Additional information:
Equilibrium is when the forward reaction rate equals the reverse reaction rate. At equilibrium, both reactant and product levels are stable.
The concentrations are usually expressed in molarity, which has units of $\dfrac{{{\text{mol}}}}{{\text{l}}}$.

Note: There is some important thing to remember when calculating the ${{\text{K}}_{\text{c}}}$;
$K_c$ is a constant for a specific reaction at a specific temperature. With change in the temperature of a reaction, then $K_c$ also changes. Pure solids and pure liquids, including solvents, are not included in the equilibrium expression. The reaction must be balanced with the coefficients written as the lowest possible integer values in order to get the correct value for $K_c$.