Equilibria, redox and acids

1. What is true at dynamic equilibrium?

1. The reaction has stopped
2. The forward and backward reaction rates are equal
3. Only the forward reaction occurs
4. Concentrations of reactants and products are always equal

Show answer

B: The forward and backward reaction rates are equal. Both reactions continue, but at matching rates.

2. What does a large value of Kc indicate?

1. Equilibrium favours the reactants
2. Equilibrium favours the products
3. The reaction is very slow
4. No reaction occurs

Show answer

B: Equilibrium favours the products. A larger numerator (products) than denominator (reactants) gives a large Kc.

3. What is the only factor that changes the value of Kc?

1. Concentration
2. Pressure
3. Temperature
4. A catalyst

Show answer

C: Temperature. Kc is fixed at a given temperature.

4. Does adding a catalyst change the position of equilibrium?

1. Yes, it shifts towards products
2. Yes, it shifts towards reactants
3. No, it only affects the rate of reaching equilibrium
4. It stops the reaction

Show answer

C: No, it only affects the rate of reaching equilibrium. A catalyst speeds up both forward and backward reactions equally.

5. Why must concentrations be at equilibrium to calculate Kc?

1. Kc can be calculated at any point in the reaction
2. The Kc expression only applies once concentrations are no longer changing
3. Kc doesn't depend on concentration
4. Equilibrium isn't needed

Show answer

B: The Kc expression only applies once concentrations are no longer changing. Before equilibrium, concentrations are still changing, so the ratio isn't constant.

6. A reaction has Kc = 500. What does this suggest?

1. The reaction barely proceeds
2. The equilibrium lies well towards the products
3. The reaction is at absolute equilibrium concentration of 500 mol/dm³
4. Kc has no meaning here

Show answer

B: The equilibrium lies well towards the products. A large Kc means a large proportion of products at equilibrium.

7. At equilibrium [H₂] = 0.20, [I₂] = 0.20 and [HI] = 0.80 mol dm⁻³. What is Kc for H₂ + I₂ ⇌ 2HI?

1. 4.0
2. 0.64
3. 16
4. 0.0625

Show answer

C: 16. Kc = 0.80² ÷ (0.20 × 0.20) = 0.64 ÷ 0.040 = 16, with no units as they cancel. Forgetting to square [HI] gives 4.0.

8. Why does raising the pressure shift N₂ + 3H₂ ⇌ 2NH₃ to the right?

1. It speeds up only the reverse reaction
2. More gas moles on the right lower the pressure
3. Fewer gas moles on the right lower the pressure
4. It increases the value of Kc

Show answer

C: Fewer gas moles on the right lower the pressure. The right has 2 mol of gas against 4 mol on the left, so shifting right reduces the pressure and counteracts the change. Kc is unchanged.

9. N₂ + 3H₂ ⇌ 2NH₃ has ΔH = −92 kJ mol⁻¹. What happens when temperature is raised?

1. Kc is unchanged and equilibrium shifts left
2. Kc increases and equilibrium shifts right
3. Kc decreases and equilibrium shifts left
4. Kc decreases and equilibrium shifts right

Show answer

C: Kc decreases and equilibrium shifts left. Raising temperature favours the endothermic reverse reaction, so equilibrium shifts left and Kc falls.

10. Why is a moderate temperature (about 450 °C) used in the Haber process?

1. Compromise between reasonable rate and acceptable yield
2. It gives the highest possible Kc
3. It prevents the catalyst being used up
4. It maximises both yield and rate

Show answer

A: Compromise between reasonable rate and acceptable yield. Low temperature favours yield (exothermic) but is too slow; high temperature is fast but gives low yield. About 450 °C is a compromise.