For the reaction
${N_2} + 3{H_2}\underset {} \leftrightarrows 2N{H_3}$
And $\dfrac{1}{2}{N_2} + \dfrac{3}{2}{H_2}\underset{{}}{\overset{{}}{\longleftrightarrow}}N{H_3}$
Write down the expression for equilibrium constant ${K_c}$ and $K_c'$. How ${K_c}$is related to $K_c'$?

Hint: Equilibrium constant expresses the relationship between products and reactants of a reaction at equilibrium. Equilibrium is a state of rest or balance due to equal action and opposing forces. There exist three types of equilibrium: stable, unstable and neutral equilibrium.
Formula used: ${K_c} = \dfrac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}}$
Where ${K_c}$ is equilibrium constant, $\left[ {N{H_3}} \right]$ is concentration of $N{H_3}$ (product), $\left[ {{N_2}} \right]$is concentration of ${N_2}$ (reactant), $\left[ {{H_2}} \right]$is concentration of ${H_2}$(reactant).

Complete step by step answer:
We know equilibrium constant expresses the relationship between products and reactants of a reaction at equilibrium. Let’s find an equilibrium constant for reaction ${N_2} + 3{H_2}\underset {} \leftrightarrows 2N{H_3}$.
In the given reaction observing stoichiometric coefficients we found that initial number of moles of ${N_2}$ are $1$, ${H_2}$ are $3$ and $N{H_3}$ are $0$. At equilibrium let concentration of $N{H_3}$ be $x$ but coefficient of $N{H_3}$ is $2$ so equilibrium concentration of $N{H_3}$ will be $2x$. Per mole consumption of reactants will be $x$. So according to stoichiometric coefficients concentration at equilibrium will be $1 - x$ for ${N_2}$ and $3 - 3x$ for ${H_2}$. Equilibrium constant is calculated as the ratio of concentration of products (raise to the power of stoichiometric coefficient) to the concentration of reactants (each reactant raises to the power of stoichiometric coefficient).
Equilibrium constant for this reaction is ${K_c}$ which will be equal to:
${K_c} = \dfrac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}}$
\[{K_c} = \dfrac{{{{\left( {2x} \right)}^2}}}{{\left( {1 - x} \right){{\left( {3 - 3x} \right)}^3}}}\] (equation a)
Similarly for reaction $\dfrac{1}{2}{N_2} + \dfrac{3}{2}{H_2}\underset{{}}{\overset{{}}{\longleftrightarrow}}N{H_3}$
In the given reaction observing stoichiometric coefficients we found that initial number of moles of ${N_2}$ are $\dfrac{1}{2}$, ${H_2}$ are $\dfrac{3}{2}$ and $N{H_3}$ are $0$. At equilibrium let concentration of $N{H_3}$ be $x$. Per mole consumption of reactants will be $x$. So according to stoichiometric coefficients concentration at equilibrium will be $\dfrac{1}{2}\left( {1 - x} \right)$ for ${N_2}$ and $\dfrac{3}{2}\left( {1 - x} \right)$ for ${H_2}$. Equilibrium constant is calculated as the ratio of concentration of products (raise to the power of stoichiometric coefficient) to the concentration of reactants (each reactant raises to the power of stoichiometric coefficient).
Equilibrium constant for this reaction is $K_c'$ which will be equal to:
$K_c' = \dfrac{{\left[ {N{H_3}} \right]}}{{{{\left[ {{N_2}} \right]}^{\dfrac{1}{2}}}{{\left[ {{H_2}} \right]}^{\dfrac{3}{2}}}}}$
\[K_c' = \dfrac{{\left( {2x} \right)}}{{{{\left( {1 - x} \right)}^{\dfrac{1}{2}}}{{\left( {3 - 3x} \right)}^{\dfrac{3}{2}}}}}\]
If we take square both sides it will be:
\[K_c' = \dfrac{{{{\left( {2x} \right)}^2}}}{{\left( {1 - x} \right){{\left( {3 - 3x} \right)}^3}}}\]
If we compare this with equation a we will find that
${K_c} = {\left( {K_c'} \right)^2}$
So this is the relationship between ${K_c}$ and $K_c'$.

Note:
If the value of equilibrium constant is greater than one then reaction is more in the forward direction and if the value of equilibrium constant is less than one this means reaction is more in the backward direction.