Question: Given \[{{K}_{c}}\] for \[PC{{l}_{5}}\overset{{}}{\leftrightarrows}PC{{l}_{3}}+C{{l}_{2}}\] is 13.7 ...

Question

Given ${{K}_{c}}$ for $PC{{l}_{5}}\overset{{}}{\leftrightarrows}PC{{l}_{3}}+C{{l}_{2}}$ is 13.7 $\dfrac{mol}{litre}$ at 546K. Calculate the pressure developed in a 10 litre box in equilibrium if the reactant concentration is 1 mole.

Answer 1

Solution

The value of ${{K}_{c}}$ is given by the division of concentration of the product to the concentration of reactants. After finding ${{K}_{c}}$ , the value of pressure can be easily found out by the formula $PV=nRT$ , as we have the values of all necessary terms.

Complete answer:
The pressure inside the container will be 8.93 bar, let’s see how.
First we need to find the value of ${{K}_{c}}$ , and to find that we have to find the concentrations of reactants and products at equilibrium.
The initial concentration of reactant will be 0.1 and the initial concentration of products $PC{{l}_{5}}$ and $C{{l}_{2}}$ will be zero.
Now, let us suppose that the equilibrium concentrations of the products will be the same as ‘x’, therefore the equilibrium concentration of the reactant would be $0.1 – x$ .
Now, as per the formula of ${{K}_{c}}$ , ${{K}_{c}}=\dfrac{\left[ PC{{l}_{3}} \right]\left[ C{{l}_{2}} \right]}{\left[ PC{{l}_{5}} \right]}$
Putting the value in the equation, ${{K}_{c}}=\dfrac{x\times x}{0.1-x}$
Now, we have been given the value of ${{K}_{c}}$ as 13.7, replacing that value in the above equation we get,
$\dfrac{x\times x}{0.1-x}=13.7$
Therefore, by making $x$ as the subject we get the value of $x$ as $0.0993$ .
Now, to find the total number of moles, $n = 10 (0.1 + x)$
$= 10 (0.1 + 0.0993)$
$= 1.993.$
As we have the values of terms volume, number of moles, universal gas constant and temperature, let us put these in $PV=nRT$ to get the value of pressure.
Therefore, $P=\dfrac{nRT}{V}$
$P=\dfrac{1.993\times 0.0821\times 546}{10}$
$P=8.93$
Therefore, the value of pressure is 8.93 bar.

Note:
The value of pressure could be easily found out, but here the value of the number of moles was not given, thus we had to find the number of moles through the concentration given at equilibrium. The value of the universal gas constant is taken in the unit of temperature as kelvin.