Question

Given 1.6 mole of $PC{{l}_{5}}$(g) is placed in a 4 dm closed vessel. When the temperature is raised to 500 K, it decomposes and at equilibrium 1.2 mole of $PC{{l}_{5}}$(g) remains. What is the ${{K}_{c}}$ value for the decomposition of $PC{{l}_{5}}$(g) to $PC{{l}_{3}}$ (g) and $C{{l}_{2}}$(g) at 5OOK?
A. 0.013
B. 0.050
C. 0.033
D. 0.067
E. 0.045

Answer 1

Hint: Try to write the equation and find the concentrations of the reactants and products at initial time and at the time equilibrium is achieved. Then find the equilibrium constant by substituting the values.

Complete answer:
In order to answer our question, we need to learn some things about the equilibrium constant ${{K}_{c}}$and the law of chemical equilibrium. The law of chemical equilibrium is a mathematical expression which can be derived by the application of the law of mass action to a reversible reaction. Consider the following reaction:$A+B\underset{{}}{\leftrightarrows}C+D$.
Now, rate of forward reaction$={{K}_{1}}[A][B]$
Rate of backward reaction=${{K}_{2}}[C][D]$
A,B,C and D are the active masses of the reactants and products and ${{K}_{1}},{{K}_{2}}$ are the rate constants. Now, at equilibrium, the rate of forward reaction is equal to the rate of backward reaction. So, we have
$\begin{align}
  & {{K}_{1}}[A][B]={{K}_{2}}[C][D] \\
 & or\dfrac{{{K}_{1}}}{{{K}_{2}}}={{K}_{c}}=\dfrac{[C][D]}{[A][B]} \\
\end{align}$
This is called the equilibrium constant for the reaction. However, the stoichiometric coefficients are raised to the power of active masses. Now Let us see our question and write the reactions for time t=0 and at time=t’(equilibrium)
$\begin{align}
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,PC{{l}_{5}}(g)\xrightarrow{equilibrium}PC{{l}_{3}}(g)+C{{l}_{2}}(g) \\
 & at\,t=0\,\,\,\,\,\,\,\,1.6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
 & at\,t=t'\,\,\,\,\,\,\,1.2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.4 \\
\end{align}$
Now, let us find the effective concentrations of the three gases in the equilibrium, so that we can calculate their equilibrium constants:
$\begin{align}
  & PC{{l}_{5}}=\dfrac{1.6}{4}mol/d{{m}^{3}} \\
 & PC{{l}_{3}}=\dfrac{0.4}{4}mol/d{{m}^{3}} \\
 & C{{l}_{2}}=\dfrac{0.4}{4}mol/d{{m}^{3}} \\
\end{align}$
So, the respective concentrations are 0.4, 0.1 and 0.1 respectively. Now, with the help of these values, we will try to find the equilibrium constant:
So, the equilibrium constant is:
${{K}_{c}}=\dfrac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}=\dfrac{0.1\times 0.1}{0.4}=0.025$
We obtain the value of equilibrium constant to be 0.0025, however answer options do not match, so the closest answer we have is 0.033.

So, our correct answer for this question will be option C.

Note:
Equilibrium constant is the same for a particular reaction only if the temperature is constant. In case the temperature increases, the rate of forward reaction may increase or decrease, based on whether the reaction is exothermic or endothermic, so the ${{K}_{c}}$ will also change.