Question: Pure \[PC{l_5}\] is introduced into an evacuated chamber and comes to equilibrium at \[{247^o}C\] an...

Question

Pure $PC{l_5}$ is introduced into an evacuated chamber and comes to equilibrium at ${247^o}C$ and $2.0$ at. The equilibrium gaseous mixture contains $40\%$ chlorine by volume
Calculate ${K_p}$ for ${247^o}C$ for the reaction.
$PC{l_5}(g)\underset {} \leftrightarrows PC{l_3}(g) + C{l_2}(g)$

Answer 1

Solution

${K_p}$ can be defined as the equilibrium constant which is used to define the equilibrium of the reaction. ${K_p}$ Is an equilibrium constant written with respect to the atmospheric pressure? There is also another term related to the equilibrium constant ${K_c}$ , which describes the equilibrium constant with respect to the concentrations of the reacting species.

Complete answer:
In order to solve this question, we have to apply the laws of stoichiometry too.
Here, as we know that the ${K_p}$ is an equilibrium constant written with respect to the atmospheric pressure.
So, now let’s solve this question by keeping in mind all the values that are given to us in the question.
So let’s proceed:
Here we are given that the gaseous mixture contains $40\%$ $PC{l_3}$ and $40\%$ $C{l_2}$ . And also we know from the equation that they both are produced in the ratio of $1:1$ .
Thus, through calculations, we can say that the $PC{l_5}$ is $20\%$
Also we know, for ideal gas moll $\%$ $= vol\%$
Now, solving the question further we will get:
${P_{C{l_2}}} = {P_{C{l_3}}}$
$2 \times 0.40 = > 0.80atm$
${P_{C{l_5}}} = 2 \times 0.2 = > 0.40atm$
So, now calculating the ${K_p}$ , we will get:
${K_p} = \dfrac{{{P_{PC{l_3}.{P_{C{l_2}}}}}}}{{{P_{PC{l_5}}}}}$
$= \dfrac{{0.80 \times 0.80}}{{0.40}}$
$= 1.6atm$
Hence, the desired ${K_p}$ is $1.6atm$

Therefore, on the basis of above calculations, the correct answer is option C.

Note:
The magnitude of the equilibrium concept is used to depict the extent of the rate of reaction. This means, if the equilibrium constant is greater, it means that the product formation is favoured, and the reaction is moving forward at a good speed. However, the equilibrium constant is affected by a few factors like temperature.