# #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?

##### 1 Answer

#### Answer:

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.

#"PCl"_ (5(g)) rightleftharpoons "PCl"_ (3(g)) + "Cl"_ (2(g))#

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

Since the reaction takes place in a *interchangeably*.

So, let's assume that you start with **does not dissociate**, which implies that

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

#6/10 * x quad "M" = (3x)/5 quad "M"#

Consequently, you can say that the reaction will produce

#["PCl"_3] = (6x)/10 quad "M" " "# and#" " ["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 notdissociate, i.e.#40%# of what you started with.

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

#K_c = (["PCl"_3] * ["Cl"_2])/(["PCl"_5])#

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

#K_c = ( (3x)/5 * (3x)/5)/((2x)/5)#

#K_c = (3x)/5 * (3color(red)(cancel(color(black)(x))))/color(red)(cancel(color(black)(5))) * color(red)(cancel(color(black)(5)))/(2color(red)(cancel(color(black)(x))))#

#K_c = 9/10 * x#

So if you know the initial concentration of the reactant, you can plug that for