#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?

Answer 1

Here's what I got.

First off, it's important to note that since you don't know the initial phosphorus pentachloride concentration, you are unable to provide a numerical solution in this case.

Therefore, you can, at most, give an expression for the reaction's equilibrium constant that depends on the reactant's initial concentration or mole count.

Write the chemical equation that describes this equilibrium reaction in a balanced manner first.

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

It is noteworthy that a dissociation of one mole of phosphorus pentachloride yields one mole of phosphorus trichloride and one mole of chlorine gas.

Since the reaction takes place in a #"1-L"# vessel, you can treat the number of moles and the molar concentration interchangeably.
So, let's assume that you start with #x# #"M"# of phosphorus pentachloride. At #300^@"C"#, you know that #40%# of this initial concentration does not dissociate, which implies that #60%# does.

This indicates that the phosphorus pentachloride concentration that undergoes reaction is equivalent to

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

As a result, you could say that the response will result in

#["PCl"_3] = (6x)/10 quad "M" " "# and #" " ["Cl"_2] = (6x)/10 quad "M"#

Phosphorus pentachloride's equilibrium concentration is going to 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.

The equilibrium constant for this reaction is defined as follows:

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

This will equal—I'll write the equilibrium constant's expression without any additional units—in your situation!

#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 #x# and find the value of the equilibrium constant.
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Answer 2

[ K_c = \dfrac{[\text{Products}]}{[\text{Reactants}]} ]

[ K_c = \dfrac{[\text{PCl}_3]}{[\text{PCl}_5]} ]

[ K_c = \dfrac{0.6}{0.4} = 1.5 ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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