Question

For the equilibrium $S{O_2}C{l_2}\left( g \right) \rightleftharpoons S{O_2}\left( g \right) + C{l_2}\left( g \right)$, what is the temperature at which $\dfrac{{{K_p}({\text{atm}})}}{{{K_c}\left( {\text{M}} \right)}} = 3$
A.$0.027{\text{ K}}$
B.$0.36{\text{ K}}$
C.$36.54{\text{ K}}$
D.273 K

Answer 1

$S{O_2}C{l_2}$ can be named as sulfuryl chloride. Sulfuryl chloride is a source of chlorine and it can be used to treat wool to prevent shrinking. ${K_p}$and ${K_c}$ is related using the formula given below.
Formula Used:
${K_p} = {K_c}{\left( {RT} \right)^{\Delta ng}}$
where R is the universal gas constant and T is the temperature and $\Delta ng$ is the difference between the number of the moles of gases in the product side and the number of the moles of gases in the reactant side.

Complete step by step answer:
The state in which the rate of the forward reaction is the same as the rate of the backward reaction, is called chemical equilibrium. In chemical equilibrium, there is no net change of reactant and product concentration. The other name of chemical equilibrium is dynamic equilibrium.
${K_p}$ is the equilibrium constant used to express the partial pressure of the reactant gases and the product gases. ${K_c}$ is the equilibrium constant used to express the molarity.
The given reaction is as follows,
$S{O_2}C{l_2}\left( g \right) \rightleftharpoons S{O_2}\left( g \right) + C{l_2}\left( g \right)$
The reactant here is sulfuryl chloride gas and the products are sulphur dioxide gas and chlorine gas. Hence the number of moles of gases are two in the product side and one in the reactant side. So, the value of $\Delta ng$ for the given reaction will be two minus one and that will be equal to one.
${K_p} = {K_c}{\left( {RT} \right)^{\Delta ng}}$
Hence for the above-mentioned reaction,
${K_p} = {K_c}{\left( {RT} \right)^1}$
Hence, $RT = \dfrac{{{K_p}}}{{{K_c}}}$
The value of $\dfrac{{{K_p}}}{{{K_c}}}$ is given as 3 and R has the fixed value of $0.0821$.
Therefore, $T = \dfrac{3}{{0.0821}} = 36.54{\text{ K}}$
Hence, the temperature will be $36.54{\text{ K}}$.

Therefore, the correct answer is option C.
Note:
The value of the universal gas constant depends on the units applied. It is equal to $8.31{\text{ J/K mol}}$ or $1.98{\text{ cal/K mol}}$ or $0.082{\text{ L at}}{{\text{m}}^3}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$