ΔG^o = −RT lnK

For a reaction to be spontaneous Gibb's free energy must be negative. This means that the equilibrium constant, K, must be greater than one (Natural log, ln, of a number less than 1 will be negative).

ΔG^o will be negative and K_c > 1 is therefore the correct answer.

You should know that Gibb's free energy, ΔG^o, is always negative for a spontaneous reaction, so we might expect the position of equilibrium to correspond to the minimum value of Gibb's free energy.

The total entropy of a system is inversely related to the Gibb's free energy, so when ΔG^o is at a minimum, ΔS_total, is at a maximum.

You need to learn that for any reaction at equilibrium: The position of equilibrium corresponds to a maximum value of entropy and a minimum value of Gibb's free energy.

Thus, Entropy is at a maximum and Gibb's free energy is at a minimum is the correct answer.

ΔG^o = −RTlnK

ΔG^o is negative, so K must be greater than one (Natural log, ln, of a value less than one will give a positive number, which would mean that ΔG^o would be positive).

And if K > 1, then the products must be favoured over the reactants (since products are on the top of the Kc expression).

The correct answer is therefore K_c > 1 and the equilibrium favours the products.

Using the initial-change-equilibrum (ICE) moles table (or RICE if you include the mole ratio):

The non-bold information is given in the question.

The change in concentration (which is the same as the change in number of moles in this example, since the volume terms will all cancel) for I₂ and Br₂ is −0.20.

The mole ratio for I₂ or Br₂ to IBr is 1:2, so 0.20 moles used up will produced +0.40 moles of IBr.

\(K_c = {[IBr]^2 \over [I_2][Br_2]}\)

\(K_c = {0.40^2 \over 0.20 \times0.20}\)

\(K_c = {0.16 \over 0.04}\)= 4

The correct answer is 4.0

Using the initial-change-equilibrum (ICE) moles table (or RICE if you include the mole ratio):

The non-bold information is given in the question.

The total volume is 1.0 dm³, so the moles values will also be the concentration values.

We have initial and equilibrium moles of N₂ so the 1st step is to calculate the change in N₂: 1.0 to 0.9 is a change of −0.1

If we have the change in N₂ we can then calculate the change in H₂ (2nd): The mole ratio is 1:3, so the change in H₂ will be −0.3 and [H₂] = 0.7.

Lastly (3rd) we can calculate the change in moles of NH₃. The mole ratio of N₂ : NH₃ is 1:2, so the change in NH₃ will be +0.2 and [NH₃] = 1.2.

Therefore the correct answer is [H₂] = 0.7 and [NH₃] = 1.2

Using the equilibrium expression:

\(K_c = {[W][Z]^2 \over [X]^2[Y]}\)

\(K_c = {1 \times4^2 \over 2 \times2^2}\)

\(K_c = {16 \over 8}\)= 2

The correct answer is 2

As the mole ratio for the equilibrium is 1 : 1 : 1 : 1, the moles of A and the moles of B that are used up, will be equal to the moles of C and moles of D that are produced. Hence, the relative concentrations of A, B, C and D at equilibrium are 1−\(x\), 1−\(x\), \(x\) and \(x\) respectively.

Using the equilibrium expression:

\(K_c = {[C][D] \over [A][B]}\)

\(K_c = {x.x \over (1-x)(1-x)}\)

\(K_c = {x^2 \over (1-x)^2}\)

Which is therefore the correct answer.

Using the equilibrium expression:

\(K_c = {[L]^2[M] \over [J][K]}\)

\(K_c = {4^2 \times2 \over 4 \times1}\)

\(K_c = {32 \over 4}\)= 8 at 300K

But 600K is higher than 300K, and at higher temperature the equilibrium will favour the reverse (endothermic) direction which will decrease the value of the equilibrium constant.

The correct answer is therefore K_c < 8