IB Chemistry Quiz: Apply Extent Of Chemical Change

What this quiz covers

This quiz focuses on Apply Extent Of Chemical Change, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The forward reaction is exothermic and K_c decreases with temperature.
2. The forward reaction is endothermic and K_c increases with temperature. (correct answer)
3. The forward reaction is exothermic and K_c increases with temperature.
4. The forward reaction is endothermic and K_c decreases with temperature.

Question 2

1. K_c = $\frac{[\text{Ca}^{2+}][\text{H}_2\text{O}][\text{CO}_2]}{[\text{CaCO}_3][\text{H}^+]^2}$
2. K_c = $\frac{[\text{Ca}^{2+}][\text{CO}_2]}{[\text{H}^+]^2}$ (correct answer)
3. K_c = $\frac{[\text{Ca}^{2+}]}{[\text{H}^+]^2}$
4. K_c = $\frac{[\text{Ca}^{2+}][\text{CO}_2]}{[\text{CaCO}_3][\text{H}^+]^2}$

Question 3

1. The system is at equilibrium.
2. The reaction quotient Q_c > K_c, so the equilibrium will shift to the left. (correct answer)
3. The reaction quotient Q_c < K_c, so the equilibrium will shift to the right.
4. The reaction quotient Q_c > K_c, so the equilibrium will shift to the right.

Question 4

1. 0.17 (correct answer)
2. 0.13
3. 0.33
4. 6.0

Question 5

1. The equilibrium shifts right and [O₂] decreases.
2. The equilibrium shifts left and [O₂] increases.
3. The equilibrium is not affected and [O₂] remains constant. (correct answer)
4. The equilibrium shifts left and [O₂] remains constant.

Question 6

1. 0.39 mol dm⁻³
2. 0.78 mol dm⁻³ (correct answer)
3. 0.44 mol dm⁻³
4. 0.11 mol dm⁻³

Question 7

1. 0.25
2. 16.0
3. -8.0
4. 0.0625 (correct answer)

Question 8

1. Equilibrium shifts to the left and solubility decreases.
2. Equilibrium shifts to the right and solubility increases. (correct answer)
3. Equilibrium shifts to the right but solubility decreases.
4. Equilibrium is unaffected as HCl is not part of the equilibrium expression.

Question 9

1. The system is at equilibrium.
2. The rate of the forward reaction is greater than the rate of the reverse reaction.
3. The rate of the reverse reaction is greater than the rate of the forward reaction. (correct answer)
4. The equilibrium constant will increase to 8.0 to match the current concentrations.

Question 10

1. The solution turns blue because the equilibrium shifts to the right. (correct answer)
2. The solution turns blue because the equilibrium shifts to the left.
3. The solution becomes a more intense pink because the equilibrium shifts to the left.
4. There is no colour change because water is a product.

Question 11

1. 0.050
2. 0.16
3. 0.20 (correct answer)
4. 0.50

Question 12

1. The forward reaction is very slow at 400 °C.
2. The reaction is exothermic.
3. At equilibrium, the concentration of NH₃ is low compared to the concentrations of N₂ and H₂. (correct answer)
4. Increasing the pressure will not increase the yield of ammonia.

Question 13

1. Both forward and reverse rates decrease, and the equilibrium shifts to the left. (correct answer)
2. Both forward and reverse rates increase, and the equilibrium shifts to the right.
3. Both forward and reverse rates decrease, and the equilibrium shifts to the right.
4. The forward rate decreases, the reverse rate increases, and the equilibrium shifts to the left.

Question 14

1. pH = 9.42 because the added base converts some $\text{NH}_4^+$ to $\text{NH}_3$, shifting the buffer composition favorably
2. pH = 9.65 because the strong base addition overwhelms the buffer capacity and significantly raises the pH
3. pH = 9.42 because the Henderson-Hasselbalch equation accounts for the stoichiometric changes in buffer components
4. pH = 9.77 because the added $\text{OH}^-$ increases both the $\text{NH}_3$ concentration and solution basicity directly (correct answer)

Question 15

1. Forward direction because $Q_p = 0.033 < K_p$, indicating insufficient product formation at equilibrium (correct answer)
2. Reverse direction because $Q_p = 1.33 > K_p$, indicating excess product concentration relative to equilibrium
3. Forward direction because $Q_p = 0.67 < K_p$, indicating the system needs more product formation
4. Reverse direction because $Q_p = 2.25 > K_p$, indicating products exceed their equilibrium concentrations

Question 16

1. pH = 4.57 with reduced buffer capacity because the unequal concentrations create an asymmetric buffering response
2. pH = 4.92 with enhanced buffer capacity because the excess acid provides additional buffering against base addition
3. pH = 4.57 with normal buffer capacity because the Henderson-Hasselbalch equation compensates for concentration differences automatically
4. pH = 4.56 with reduced buffer capacity because the 3:2 ratio decreases the minimum concentration component (correct answer)

Question 17

1. The compression shifts equilibrium left, resulting in 0.156 mol $\text{PCl}_3$ and 0.344 mol $\text{PCl}_5$ at new equilibrium
2. The compression shifts equilibrium left, resulting in 0.111 mol $\text{PCl}_3$ and 0.389 mol $\text{PCl}_5$ at new equilibrium (correct answer)
3. The compression shifts equilibrium right, resulting in 0.201 mol $\text{PCl}_3$ and 0.299 mol $\text{PCl}_5$ at new equilibrium
4. The compression has no effect because $K_p$ remains constant, maintaining 0.178 mol $\text{PCl}_3$ and 0.322 mol $\text{PCl}_5$

Question 18

1. pH = 1.70 because the HCl completely dissociates without buffering capacity present (correct answer)
2. pH = 2.30 because partial neutralization occurs even without buffer components
3. pH = 3.15 because some hydrolysis of chloride ions provides slight buffering
4. pH = 4.20 because the pH change should be similar to the buffered solution

Question 19

1. A precipitate forms because $Q_{sp} = 6.0 \times 10^{-8} > K_{sp} = 1.69 \times 10^{-14}$
2. No precipitate forms because $Q_{sp} = 3.0 \times 10^{-8} < K_{sp} = 1.69 \times 10^{-14}$
3. A precipitate forms because $Q_{sp} = 3.0 \times 10^{-8} > K_{sp} = 1.69 \times 10^{-14}$ (correct answer)
4. No precipitate forms because $Q_{sp} = 1.5 \times 10^{-8} < K_{sp} = 1.69 \times 10^{-14}$

Question 20

1. Additional $\text{Mg(OH)}_2$ precipitates because the common ion effect increases $Q_{sp}$ above $K_{sp}$ significantly
2. Additional $\text{Mg(OH)}_2$ precipitates because added $\text{Mg}^{2+}$ shifts equilibrium toward solid formation (correct answer)
3. Some $\text{Mg(OH)}_2$ dissolves because dilution decreases ion concentrations below the $K_{sp}$ threshold temporarily
4. No change occurs because the system maintains equilibrium through equal dissolution and precipitation rates