Question

A quantity of $\mathrm{PCl}_{5}$ was heated in a 10 jjtre vessel at $250^{\circ} \mathrm{C} ; \mathrm{PCl}_{\text {spi }} \rightleftharpoons \mathrm{PCl}_{3(0)}+\mathrm{Cl}_{2(\text { anj }}$. At
equilibrium, the vessel contains $0.1$ mole of $\mathrm{PCl}_{\mathrm{s}_{r}} 0.20$ mole of $\mathrm{PCl}_{3}$, and $0.2$ mole of $\mathrm{Cl}_{3}$. The equilibrium constant of the reaction is:
A. $0.02$
B. $0.05$
$C. 0.04$
D. 0.025

Answer 1

Hint: If any reaction reaches the state of equilibrium, then with the help of the equilibrium constant you can find the amount of reactant with respect to the product or vice versa at the eqbm. position. The equilibrium constant (K) is defined as the ratio of the product of the concentration of product each raised to the power of their stoichiometric coefficient to the product of the concentration of reactants each raised to the power of their stoichiometric coefficient such as:
For a general reaction;
$a A+b B \rightleftharpoons c C+d D$, equilibrium constant is defined as
$\left.k_{c}=[C][D]\right]^{4} /\left[A^{0}[B]^{b}\right.$; where $[C]$ and $[D]^{d}$, square measure the molar concentration of products, $C$ and $D$ both raised to the power of their stoichiometric coefficients (number of moles) and similarly for reactant.

Complete Step by Step Answer:
For a given chemical equation,
$\mathrm{PCl}_{\text {spipl }} \rightleftharpoons \mathrm{PCl}_{3(6)}+\mathrm{Cl}_{2(\mathrm{~g})}$
$k_{\mathrm{c}}=\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right] /[\mathrm{PCl} 5]$
$\left[\mathrm{PCl}_{5}\right],\left[\mathrm{PCl}_{3}\right]$, and $\left[\mathrm{Cl}_{2}\right]$ are the concentration of $\mathrm{PCl}_{5}, \mathrm{PCl}_{3}$, and $\mathrm{Cl}_{2}$.
Now, molar concentration is defined as the ratio of the number of moles of solute to the volume of
solution in litre (10 litre as per question) such as
$\left[\mathrm{PCl}_{5}\right]=0.1 / 10=0.01\left(0.1\right.$ mole of $\left.\mathrm{PCl}_{5}\right)$
$\left[\mathrm{PCl}_{3}\right]=0.20 / 10=0.02\left(0.20\right.$ mole of $\left.\mathrm{PCl}_{3}\right)$
$[\mathrm{Cl} 2]=0.2 / 10=0.02\left(0.2\right.$ mole of $\left.\mathrm{Cl}_{2}\right\}$
Now, the value of $k_{c}$ is given as
$k_{c}=\left[\mathrm{PCl}_{3}\right]\left[\right.$ Csimilar $\left.\mathrm{L}_{5}\right]$
$k_{c}=(0.02 * 0.02) / 0.01$
$k_{c}=0.04 \mathrm{~mol} /$ litre
Thus, the correct option is $\mathrm{C}$.

Note: In this question, the concentration of compounds is not given but the mole of solutes and volume of solvent is given. Thus, we need to calculate the molar concentration of every solute and then need to put all these concentrations in the formula of ari equilibrium constant to get its value.