Question

Two moles of $PC{{l}_{5}}$were introduced in a 2-liter flask and heated at 600 K to attain equilibrium. $PC{{l}_{5}}$ was found to be 40% dissociated intoPC{{l}_{3}}$$$$C{{l}_{2}}, Calculate the value of${{K}_{c}}$.

Answer 1

The value of equilibrium constant can be calculated by dividing the concentration of the products with the concentration of the reactant. For finding the value of the concentration of each compound the calculated number of moles should be divided by the volume of the flask.

Complete step by step answer:
The product of the molar concentration of the products, each raised to the power to its stoichiometric coefficient divided by the product of the molar concentration of the reactant, each raised to the power to its stoichiometric coefficient at constant temperature is called the Equilibrium constant.
Now, according to the question, the$PC{{l}_{5}}$ dissociates into PC{{l}_{3}}$$$$C{{l}_{2}}and.
So, the equilibrium equation will be,
$PC{{l}_{5}}\rightleftharpoons PC{{l}_{3}}+C{{l}_{2}}$
The initial amount of $PC{{l}_{5}}$ taken = 2 moles ( Given in the question)
After the equilibrium has reached 40% of $PC{{l}_{5}}$ is dissociated into $PC{{l}_{3}}$and $C{{l}_{2}}$
Therefore, $PC{{l}_{5}}$ dissociated at equilibrium = $\dfrac{40}{100}\text{ x 2 = 0}\text{.8 mole}$
Therefore, the amount of$PC{{l}_{5}}$, $PC{{l}_{3}}$ and $C{{l}_{2}}$ at the equilibrium will be:
$PC{{l}_{5}}$ = 2 – 0.8 = 1.2 mole
$PC{{l}_{3}}$ = 0.8 mole
$C{{l}_{2}}$ = 0.8 mole
This is because one mole of $PC{{l}_{5}}$ on dissociation gives 1 mole of $PC{{l}_{3}}$ and 1 mole of $C{{l}_{2}}$
Since the volume of the vessel is 2 liters, therefore, the molar concentrations at equilibrium can be calculated by dividing the number of moles with the volume of the vessel.
$[PC{{l}_{5}}]=\dfrac{1.2}{2}=0.6mol\text{ }{{\text{L}}^{-1}}$
For, $[PC{{l}_{3}}]=\dfrac{0.8}{2}=0.4mol\text{ }{{\text{L}}^{-1}}$
For, $[C{{l}_{2}}]=\dfrac{0.8}{2}=0.4mol\text{ }{{\text{L}}^{-1}}$
For,
Now, applying the law of chemical equilibrium to the dissociation equilibrium, we get
${{K}_{c}}=\dfrac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}=\dfrac{0.4mol\text{ }{{\text{L}}^{-1}}\text{ x }0.4mol\text{ }{{\text{L}}^{-1}}}{0.6mol\text{ }{{\text{L}}^{-1}}}=0.267mol\text{ }{{\text{L}}^{-1}}$

Hence the value of ${{K}_{c}}$ is $0.267mol\text{ }{{\text{L}}^{-1}}$

Note: The equation should be balanced for finding the number of moles. For finding the equilibrium constant the concentration of product and reactant should be taken. If you take the number of moles the answer would be wrong. So make sure that the number of moles of reactant and product are converted into concentrations. The ${{K}_{p}}$ of the reaction can also be calculated with ${{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}$ , where R is the gas constant, T is the temperature and $\Delta n$ is difference of moles of product and reactant.