Question

40% of Pcl5 is not dissociated at 300 degree Celsius.The reaction is carried out in a flask of 1 l capacity. The value of Kc would be?

Answer 1

Answer:- Kc = 0.9

Solution:- As per the given information, 40% of $PCl_5$ is not dissociated means 60% of it is dissociated. Let's say there is 1 mol of $PCl_5$ .

Dissociation equation for $PCl_5$ is written as:

$PCl_5(g)\Leftrightarrow PCl_3(g)+Cl_2(g)$

Since, 60% dissociation takes place, the moles of each products would be 0.6 and remaining moles of the reactant would be 1 - 0.6 = 0.4

The volume of the container is 1 L so the concentration for each one of them would be same as their moles.

$[PCl_5]=0.4M$

$[PCl_3]=0.6M$

$[Cl_2]=0.6M$

Kc expression is written as:

$Kc=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

Let's plug in the concentrations in this expression and do calculations:

$Kc=\frac{(0.6)^2}{0.4}$

Kc = 0.9

Hence, the Kc for the dissociation equation of phosphorous pentachloride is 0.9.

Answer 2

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