For the reaction $2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}$, the unit of $\mathrm{kc}$
A. litre/mole
$B.$ mol litre
$C. \ \left(m e / / l^{-1} e^{-1}\right)^{-1}$
D. (litre mole $\left.{ }^{-1}\right)^{2}$

Hint: Unit of kc (equilibrium constant) can be calculated by separately putting the unit of concentration of product and concentration of reactant (mpl litre ${ }^{1}$ ) each raise to the power of their stoichiometric coefficients (number of gaseous moles) and simplifying it.

Complete Step by Step Answer:
If any reaction reach the state of equilibrium, then with the help of equilibrium constant you can find the
amount of reactant with respect to product or vice versa at the egbol. position. Equilibrium constant (K)
is defined as the ratio of product of concentration of the product each raise to the power of their
stoichiometric coefficient to the product of concentration of reactants each raise to the power of their
stoichiometric coefficient such as

For reaction,
$a \mathrm{~A}+\mathrm{bB} \rightleftharpoons \mathrm{CC}+\mathrm{dD}$, equilibrium constant is defined as
$k_{c}=[C][D]^{d} /[A]^{2}[B]^{t}$; where $[C]^{c}$ and $[D]^{d}$ are concentration of products, $C$ and $D$ both raised to the power of their
stoichiometric coefficients, $c$ and $d$ (number of moles) and similarly for reactant.

For given chemical equation,
$2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}$, Equilibrium constant will be
$k_{c}=\left[\mathrm{SO}_{3}\right]^{2} /\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]$

As unit of concentration is same for all reactant and product and is mol litre ${ }^{-1}$ or mol per litre thus;
$k_{c}=\left[\text { mol litre }{ }^{-1}\right]^{2} /\left[\text { mol litre }{ }^{-1}\right]^{2}\left[\right.$ mol litre $\left.{ }^{-1}\right] 1$
$k_{c}=[\text { mol litre }]^{I}$
$k_{i}=m o l^{1}$ litre.
Thus, the correct option is A.

Note: It is important to know that, the unit of kc is $\dfrac{g_eq}{l}$ litre inverse for all reactions just varies through the stoichiometric coefficients (number of moles) of reactant and product. As $\mathrm{kc}$ is the ratio of concentration of product to the concentration of reactant, then their stoichiometric coefficients of reactant is 5 subtracted from stoichiometric coefficients of product (delta ng) such as $k_{c}=\left(\text { mol litre }{ }^{-1}\right)^{\text {beta ng }}$
Delta $n g=$ number of moles of product $-$ number of moles of reactant
Delta $n g=2-3$
Delta ng $=-1$
Thus, $k_{c}=[\text { mol litre }]^{1}$
$k_{c}=m o l^{-1}$ litre.