Question

In a $500 \mathrm{~mL}$ capacity vessel, $\mathrm{CO}$ and $\mathrm{Cl}_{2}$ are mixed to form $\mathrm{COCl}$. At equilibrium, it contains $0.2$ mole of $\mathrm{COCl}_{2}$ and $0.1$ mole of each of $\mathrm{CO}$ and $\mathrm{Cl}_{2}$. The equilibrium constant $\mathrm{k}_{\mathrm{E}}$ for reaction, $\mathrm{CO}+\mathrm{Cl}_{2} \rightleftharpoons$ $\mathrm{COCl}_{2}$ is
A. 5
B. 10 .
C. 15
D. 20

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 $(\mathrm{K})$ is defined as the ratio of the concentration of products, each raised to the power of their stoichiometric coefficient to the product of the concentration of reactants, each raised to power their stoichiometric coefficient.

Formula Used: Forgeneral reaction,
$\mathrm{aA}+\mathrm{bB} \rightleftharpoons \mathrm{CC}+\mathrm{dD}$, equilibrium constant is defined as
$K c=[C]^{c}[D] d[A]^{2}[B]^{b}$, Where $\left.[C]\right]^{c}$ and $[D]^{d}$ are the molar concentration of products, $C$ and $D$ both raised to
the power of their stoichiometric coefficients, $c$ and $d$ (number of moles), and similarly for reactants.

Complete Step by Step Answer:
Given equation,
$$
\mathrm{CO}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{COCl}_{2}
$$
The equilibrium constant of reaction can be found, when the molar concentration of all the reactant and product spice is known. In the given question moles of reactant and product are given and also the volume of solution is given, $500 \mathrm{~mL}$ such as
$\mathrm{CO}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{COCl}_{2}$
$0.10 .10 .2 \quad$ (at equilibrium)
Molar concentration is defined as the ratio of moles of solute to the volume of solution in litre. Thus,
$\left[\mathrm{COCl}_{2}\right]=0.2 \mathrm{mp} /(0.500 \mathrm{~L} \quad$ (H is indicating molar concentration)
$[\mathrm{CO}]=0.1 \mathrm{~mol} / 0.500 \mathrm{~L}$
$\left[\mathrm{Cl}_{2}\right]=0.1 \mathrm{mpol} / 0.500 \mathrm{~L}$
Thus, equilibrium constant, $k$ is
$k_{c}=\left[\mathrm{COCl}_{2}\right] /\left[\mathrm{CO}_{\left[\mathrm{Cl}_{2}\right]}\right.$
$\mathrm{k}_{\mathrm{c}}=(0.2 \mathrm{Dol} / 0.500 \mathrm{~L}) /(0.1 \mathrm{~mol} / 0.500 \mathrm{~L})(0.1 \mathrm{Dol} / 0.500 \mathrm{~L})$
$k_{i}=10$
Thus, the correct option is B.

Note: The volume of solution is given in millilitres thus, first we need to convert millilitre into litre as given below:
The volume of solution given in poillifitre $=500 \mathrm{~mL}$
The volume of solution in lite $=500 \mathrm{ml} \times 14 / 1000 \mathrm{~m}$
$=0.500 \mathrm{~L}$