5.1.2 (e) Calculations of Kc & Kp and determination of units.

Syllabus

(e) calculations of K_c and K_p, or related quantities, including determination of units.

{Learners will not be required to solve quadratic equations}

What does this mean?

Determining Units.

We saw in (3.2.3 (f,g) The Equilibrium Constant, Kc) how to calculate K_c given the concentrations of reactants and products.

eg K_c = [SO₃]² /[O₂] x [SO₂]²

If [SO₃]=8 moldm^-3 ,[O₂] = 4 moldm^-3 , [SO₂]=2 moldm^-3 K_c = = 8²/(4 x 2²) = 64/(4 x 4) =64/16 = 4

But we didn't learn how to calculate the units of Kc.

In this case, Kc = (moldm^-3)² / moldm^-3 x (moldm^-3)²

So, the units of this Kc are 1/moldm^-3 = mol^-1dm³

But since the Kc expression changes with each equilibrium the units are not always the same.

eg Equal moles on both sides A + B ⇌ C + D, or 2A ⇌ C+D, or A + B ⇌ 2 C etc

Kc = (moldm^-3)² / (moldm^-3)²

And there would be no units.

eg 1 more product mole than reactant moles A ⇌ B + C,

I there is one more mole on the product side (right)

So there will be 1 more moldm^-3 on the bottom of the Kc expression.

i) Kc = (moldm^-3) / moldm^-3 x (moldm^-3)

And the units will be moldm^-3

eg 1 more reactant mole than product moles i) A + B ⇌ C, or ii) 2A + B ⇌ C + D etc

In both cases, there is one more mole on the reactant side (left)

i) Kc = moldm^-3 x (moldm^-3) / (moldm^-3)

or ii) Kc = moldm^-3 x moldm^-3 x (moldm^-3) /moldm^-3 x (moldm^-3)

So there will be 1 more moldm^-3 on the bottom of the Kc expression.

And the units will be mol^-1dm³

The same process can be carried out with K_p, with the only difference being that the base unit is generally kPa rather than mol^-1dm³

^{Finding K}_c

^{A typical exam question will give you information that requires you to think about BEFORE and AFTER equilibrium has been achieved.}

^{3.36 mols of A and 0.72 moles of B are mixed in a 4 dm3 reaction vessel and allowed to react.}

^{When the mixture has come to equilibrium it is found that there are 2.88 moles of B.}

A_(g) ⁺ B_(g) ⇌ C_(g)

The easiest way to deal with this is to set out your answer in a table that allows you to keep track of which number is which:

^{Initially, these are the only numbers we know.}

However, it is clear 3.36 - 2.88 =0.48 moles of A have been destroyed.

So looking at the equation we must also have destroyed 0.48 moles of B.

And we must have created 0.48 moles of C

It would be tempting but wrong to put these numbers straight into the K_c.

Remember that K_c uses [] to represent concentrations and not moles.

To turn moles into concentration we divide by the volume of the container.

So it's a good idea to draw your table with the concentration line before you do any Maths, thus reminding you of the need to carry out this step.

From time to time examiners miss out the volume of the container. In these cases the volumes will cancel out of the calculation and so are irrelevant.

Now we can deal with the K_c.

K_c = [A]x[B] /[C] = (0.68 x 0.0525)/0.12 = 0.2975 moldm^-3

Videos

Finding K_p from moles

^{10 mols of A and 4 moles of B are allowed to react.}

^{When the mixture has come to equilibrium at 200 kPa it is found that there are 3 moles of C have.}

2A_(g) ⁺ B_(g) ⇌ 2C_(g)

Looking at the A:B:C::2:1:2 ratio it should be clear that creating 3 moles of C should have destroyed 3 moles of A

But it should have destroyed half as much (1.5 moles) of B.

The total number of moles = 7 + 2.5 + 3 = 12.5

Partial Pressure (A) = Mole Fraction (A) x Total Pressure

Finally we are ready to use the K_p.

K_p = p²C /p²A x pB

K_p = (48)² / {(112)² x 40}= 0.005 kPa^-1

^Videos

Back to 5.1.2?

Click here