IBDP Chemistry - Reactivity 2.3 How far? The extent of chemical change- IB Style Questions For SL Paper 1A - FA 2025
▶️ Answer/Explanation
1. Analyze Effect of Pressure:
Moles of gas on left = 2. Moles of gas on right = \(1+1=2\).
Since moles are equal, changing pressure has no effect on the position of equilibrium. (Any “shift” is due to temperature only here, so pressure options are irrelevant distractions, or rather, just ‘no change’ is ideal, but we have to pick the best row).
2. Analyze Effect of Temperature:
\(\Delta H = +9\) kJ (Endothermic). Heat is a reactant.
\(2HI + \text{Heat} \rightleftharpoons H_2 + I_2\)
To shift to the left (reactants), we must remove the heat. We need to decrease the temperature.
3. Evaluate Options:
We need “Temperature decreases”.
Row B has “Temperature decreases”. (It also says Pressure increases, which has no effect, so the net effect is shift left).
Row C has “Temperature increases” (shifts right).
Rows A/D have “no change” or “increases” in temp? No, A/D say no change in Temp.
Only B provides a condition (Temperature decrease) that causes a left shift.
✅ Answer: (B)
▶️ Answer/Explanation
1. Meaning of \(Q\):
The reaction quotient \(Q\) is the ratio of the activities (or concentrations/pressures) of products to reactants at a given moment:
\(Q = \dfrac{\text{(products)}^{\text{coefficients}}}{\text{(reactants)}^{\text{coefficients}}}\).
A smaller value of \(Q\) means a smaller relative amount of products compared to reactants.
2. Compare the given values:
\(4.9 \times 10^{-3} = 0.0049\)
\(8.2 \times 10^{-3} = 0.0082\)
\(4.9 \times 10^{2} = 490\)
\(8.2 \times 10^{2} = 820\)
Clearly, the smallest value is \(4.9 \times 10^{-3}\).
3. Conclusion:
The smallest \(Q\) corresponds to the lowest relative amount of products.
✅ Answer: (A)
▶️ Answer/Explanation
1. Apply Le Châtelier’s principle:
The equilibrium
\(\text{Ag}^{+}(\text{aq}) + \text{Cl}^{-}(\text{aq}) \rightleftharpoons \text{AgCl}(\text{s})\)
will shift to the right if the concentration of reactants \(\text{Ag}^{+}\) or \(\text{Cl}^{-}\) is increased, or if product is removed.
2. Consider each option:
A. Removing some solid \(\text{AgCl}(\text{s})\) does not change the equilibrium position because the activity of a pure solid is constant.
B. Adding water dilutes both ions equally; the solubility effect is small and does not clearly drive the equilibrium strongly to the right in this precipitation context.
C. Adding solid \(\text{NaCl}\) increases \([\text{Cl}^{-}]\) in solution. According to Le Châtelier’s principle, the system responds by consuming extra \(\text{Cl}^{-}\) and \(\text{Ag}^{+}\) to form more \(\text{AgCl}(\text{s})\), shifting equilibrium to the right.
D. Increasing pressure mainly affects gases; here only aqueous and solid species are present, so there is essentially no effect on equilibrium position.
3. Conclusion:
Only adding solid \(\text{NaCl}\) increases \([\text{Cl}^{-}]\) and shifts the equilibrium to the right.
✅ Answer: (C)