Equilibrium (AHL)

What does this mean?

We know from Year 12 that a reaction

2W + 3X ⇌ Y + 2Z

has a predictable K_c

K_c = [Y][Z]² / [W]²[X]³

Hopefully, you'll either recall or see from the expression that the larger the K_c the more products there must be and so the higher the yield.

Examiners like to set questions that tell you some information about the number of moles that you start with, and then some information about how things have changed when it reaches equilibrium.

Then they ask you to calculate the K_c and its units.

If we put the units for concentration (moldm^-3) into the above example;

K_c = [Y][Z]² / [W]²[X]³

Kc = (moldm^-3) x (moldm^-3)² / (moldm^-3)² x (moldm^-3)³

Then cancel:

Kc = (moldm^-3) x (moldm^-3)² / (moldm^-3)² x (moldm^-3)³

Kc = (moldm^-3) / (moldm^-3)^{3 2}

Kc = 1/ (moldm^-3)²

Kc = 1/mol²dm^-6

Units are therefore mol^-2dm⁶

In practice, it's easier to cancel the [ ]

K_c = [Y][Z]² / [W]²[X]³

K_c = [Y] / [X]³²

K_c = 1 / [X]²

K_c = [X]^-2=[moldm^-3]^-2 = mol^-2dm⁶

Find the units for the following K_cs

a) K_c = [C][D]² / [A][B]

a) K_c = [C][D]² / [A][B]²

a) K_c = [C]³[D]² / [A]³[B]

a) K_c = [C] / [A]²[B]²

a) K_c = [C][D]² / [A]³

a) K_c = [C]² / [A][B]

a) K_c = [C]²[D]² / [A]³

a) Kc = [C][D] / [A][B]²

a) K_c = [C]²[D]² / [A][B]

A chemist mixes 0.9 moles of A with 0.3 moles of B in a 5 dm³ container.

At equilibrium she finds that there is 0.2 moles of C.

2A + B ⇌ C + 2D

Now we use the stoichiometry to work out how the number of moles of A, B and D at equilibrium.

The logic is that C:D is 1:2.

So for every 1 mole of C created, 2 moles of D are created.

So, since 0.2 moles of C were created there must also have been 0.4 moles of D created.

C:B is 1:1

So for every 1 mole of C created, 1 moles of B is destroyed.

So, since 0.2 moles of C were created 0.2 moles of B were destroyed.

Final moles of B = 0.3 - 0.2 = 0.1

C:A is 1:2

So for every 1 mole of C created, 2 moles of A is destroyed.

So, since 0.2 moles of C were created 0.4 moles of A were destroyed.

Final moles of A = 0.9 - 0.4 = 0.5

DON'T put these in the K_c yet.

[A] means CONCENTRATION, these are still MOLES.

Concentration = moles/ volume, and the question says the container is 5dm³

K_c for 2A + B ⇌ C + 2D = [C][D]² / [A]²[B]

Kc = 0.04 x (0.08)² / (0.1)² x 0.02 = 1.28

There are no units.

1. A chemist mixes 2.4 moles of A with 1.6 moles of B in a 4 dm³ container.

At equilibrium, she finds that there is 0.4 moles of C.

2A + B ⇌ C + 2D

a) Write an expression for K_c

K_c =

b) Complete the table and use it to calculate K_c

K_c = (don't forget to work out units)

2. A chemist mixes 2.8 moles of W with 1.0 moles of X in a 4 dm³ container.

At equilibrium she finds that there is 0.2 moles of Z.

W + 2X ⇌ Y + Z

a) Write an expression for K_c

K_c =

b) Complete the table and use it to calculate K_c

K_c = (don't forget to work out units)

3. A chemist mixes 1.4 moles of A with 0.6 moles of B in a 2 dm³ container.

At equilibrium she finds that there is 0.2 moles of C.

A + B ⇌ C + 2D

a) Write an expression for K_c

K_c =

b) Calculate K_c (make your own table)

4. A chemist mixes 2.4 moles of Q with 2.6 moles of R in a 2 dm³ container.

At equilibrium she finds that there is 0.1 moles of T.

2Q + 3R ⇌ S + 2T

a) Write an expression for K_c

K_c =

b) Calculate K_c (make your own table)

5. A chemist puts 3.4 moles of A in a 4 dm³ container.

At equilibrium she finds that there is 0.3 moles of W.

4V ⇌ W + 2X

a) Write an expression for K_c

K_c =

b) Calculate K_c (make your own table)

You'll remember that Gibbs Free Energy is used to work out whether a reaction is feasible (spontaneous).

If ΔG is negative or zero then the reaction is feasible.

The first temperature at which it is feasible is when ΔG is zero.

And we rearranged the equation ΔG = ΔH -TΔS for ΔG =0 to get T = ΔH /ΔS

We should also know enough about K_c to realise that a High K_c means the equilibrium is on the right (there are lots of products)

And a Low K_c means the equilibrium is on the left (there are lots of reactants)

You may even recall the Arrhenius equation!

K = Ae ^(-Ea/RT)

Which can be rearranged to the equation below (no one will ask you to do this!)

However, since you know that ΔG = -ve means the reaction should happen, this means that the reaction should be going forwards and therefore that K_c should be bigger than 1.

And, since you know that ΔG = +ve means the reaction shouldn't happen, this means that the reaction should be going backwards and therefore that K_c should be smaller than 1.

And, since you know that ΔG = 0 means neither reaction is favoured, this means that K_c should be 1.

Thankfully, ΔG = - RT ln(K) is in the data-book, so you don't have to learn it.

You are quite likely to be asked to calculate K_c from ΔG.

Obviously, ln K = - ΔG / RT

So, K = e ^{(-ΔG /RT)}

So, to calculate K_c at 10 K for the reaction Ni_(s) + Cl_{2 (g)} --> NiCl_2(s)

where ΔH= -305.3 kJmol^-1 and ΔS = -155.21 JK^-1mol^-1

ΔG = ΔH - TΔS

ΔG = -305.3 - 10 (-0.15521) = -303.75 kJmol^-1

-ΔG/RT = 303.75/ 8.314 x10 = 3.65

K = e ^(3.65) = 38.5

So the reaction is favoured because the equilibrium constant is above 1, but not very heavily favoured because the number is small.

1. For the reaction CH_4(g) + 2 O_2(g) → CO_2(g) + 2 H₂O_(g)

ΔH° = -802.5 kJ

ΔS° = -4 J/K

a) Calculate ΔG° (ie at 25^oC)

b) Hence calculate K_c at 25^oC

2. For the reaction P₄O_10(s) + 6 H₂O_(l) → 4 H₃PO_4(s)

ΔH° = -416 kJ

ΔS° = -209 J/K

a) Calculate ΔG° (ie at 25^oC)

b) Hence calculate K_c at 25^oC

2. For the reaction HCl_(g) + NH_3(g) → NH₄Cl_(s)

ΔH° = -176 kJ

ΔS° = -284 J/K

a) Calculate ΔG°

b) Hence calculate K_c