Question

When $PC{l_5}$ ​ is heated in a closed container at $575K$, the total pressure at equilibrium is found to be $1$ atm and partial pressure of $C{l_2}\;$ is found to be $0.324\;$atm. Calculate the equilibrium constant (${K_P}$ ​) for the decomposition reaction.
A. $0.35$
B. $0.426$
C. $0.298$
D. None of these

Answer 1

$PC{l_5}$ at equilibrium gives $PC{l_3}$ along with $C{l_2}$, using total pressure at equilibrium and initial pressure can help in finding the complete answer. Also some calculations are required to formulate the end result among the options.

Complete step-by-step answer: Given reaction:
$\Rightarrow$ $PC{l_5}\left( g \right) \rightleftharpoons PC{l_3}\left( g \right) + C{l_2}\left( g \right)$
Initial $P\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0$
Pressure
At eq. $P-p\,\,\,\,\,\,\,\,\,\,\,\,\,\,p\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p$
Total pressure at equilibrium is 1 atm.
$\Rightarrow P - p + p + p = 1$
$\Rightarrow P + p = 1$ $\to$ $(1)$
& partial pressure of C{l_2}$$$$ = 0.324
$p = 0.324$ $\to$ (2)
After using $AB \to A + B$ & (2)$$$ \Rightarrow $$$P = 0.676atm
Now, ${K_P} = \dfrac{{pC{l_2}.PC{l_3}}}{{PPC{l_5}}}$
Where,$pC{l_2}$,$PC{l_3}$,$PpC{l_5}$are partial pressure of $C{l_2}$,$PC{l_3}$,$PC{l_5}$at equilibrium.
$\Rightarrow {K_P} = \dfrac{{0.324 \times 0.324}}{{0.676 - 0.324}}$
After solving,
$\Rightarrow {K_P} = 0.298$

ADDITIONAL INFORMATION:
${K_P}$ is known as the equilibrium partial pressure whereas ${K_C}$ is known as equilibrium constant $K$ for concentration.
The equilibrium constant, ${K_C}$ is termed as the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.
The equilibrium constant, ${K_P}$ is obtained from the partial pressures of a chemical reaction equation. It is used to tell the relationship b/w product pressures and reactant pressures. It is a unit-less no., however it relates the pressures.
A state of rest or balance because of the same and equal action of opposing forces. Equal balance b/w any powers, influences, etc.; equality of effect either mental or emotional balance.
A decomposition reaction is known as a reaction in which a compound breaks into 2 or more simpler substances. The general form of this reaction is: $AB \to A + B$ . Most decomposition reactions needs an input of energy in the manner of heat, light, or electricity.

Hence, Option C is correct.

Note: The reaction should be carefully written first then initial pressure and partial pressure must be noted carefully. Using all the given details, along with the calculations, the answer can be calculated.