# 40% of "PCl"_5 is not dissociated at 300^@"C". The reaction is carried out in a flask of "1-L" capacity. The value of K_c would be?

Mar 16, 2018

Here's what I got.

#### Explanation:

For starters, it's worth pointing out that you can't provide a numerical solution here because you don't know the initial concentration of phosphorus pentachloride.

So at best, you can provide an expression for the equilibrium constant of the reaction that depends on the initial concentration or on the initial number of moles of the reactant.

Start by writing the balanced chemical equation that describes this equilibrium reaction.

${\text{PCl"_ (5(g)) rightleftharpoons "PCl"_ (3(g)) + "Cl}}_{2 \left(g\right)}$

Notice that every mole of phosphorus pentachloride that dissociates produced $1$ mole of phosphorus trichloride and $1$ mole of chlorine gas.

Since the reaction takes place in a $\text{1-L}$ vessel, you can treat the number of moles and the molar concentration interchangeably.

So, let's assume that you start with $x$ $\text{M}$ of phosphorus pentachloride. At ${300}^{\circ} \text{C}$, you know that 40% of this initial concentration does not dissociate, which implies that 60% does.

This means that the concentration of phosphorus pentachloride that reacts is equal to

$\frac{6}{10} \cdot x \quad \text{M" = (3x)/5 quad "M}$

Consequently, you can say that the reaction will produce

["PCl"_3] = (6x)/10 quad "M" " " and $\text{ " ["Cl"_2] = (6x)/10 quad "M}$

The equilibrium concentration of phosphorus pentachloride will be

["PCl"_5] = (1 - 6/10) x quad "M"

["PCl"_5] = (2x)/5 quad "M"

This represents the concentration of phosphorus pentachloride that does not dissociate, i.e. 40% of what you started with.

By definition, the equilibrium constant for this reaction takes the form

${K}_{c} = \left(\left[{\text{PCl"_3] * ["Cl"_2])/(["PCl}}_{5}\right]\right)$

In your case, this will be equal to--I'll write the expression of the equilibrium constant without added units!

${K}_{c} = \frac{\frac{3 x}{5} \cdot \frac{3 x}{5}}{\frac{2 x}{5}}$

${K}_{c} = \frac{3 x}{5} \cdot \frac{3 \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}}} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}}}{2 \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}$

${K}_{c} = \frac{9}{10} \cdot x$

So if you know the initial concentration of the reactant, you can plug that for $x$ and find the value of the equilibrium constant.