IB DP Chemistry -Reactivity 2.3 How far? The extent of chemical change- IB Style Questions For HL Paper 1A -FA 2025

IBDP Chemistry HL Paper 1A & 1B - All Chapters

Question

Consider the equilibrium reaction:
\(\displaystyle N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\)
with \(\Delta H_r^\ominus = -46\,\text{kJ mol}^{-1}\).

Which row correctly indicates how a pressure change initially affects the equilibrium position and the comparison between \(Q\) and \(K\)?

OptionChange in
pressure		Equilibrium
shift		Relation
between \(Q\) and \(K\)
Adecreaseto the left\(Q < K\)
Bincreaseto the left\(Q > K\)
Cdecreaseto the right\(Q > K\)
Dincreaseto the right\(Q < K\)
▶️ Answer/Explanation
Detailed solution

The reaction has \(4\) moles of gas on the left and \(2\) on the right. Increasing pressure favours the side with fewer gas moles → shift **to the right**.

Right after raising the pressure, the system contains relatively too many reactants, so:

\(Q < K\)

Therefore the correct row is option D.

✅ Answer: (D)

Question

This reaction has an equilibrium constant \( K_{\mathrm{c}} = 650 \) at a certain temperature.
\[ \mathrm{NO}_2(\mathrm{g}) + \mathrm{SO}_2(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g}) + \mathrm{SO}_3(\mathrm{g}) \]
What is the equilibrium constant for the following reaction at the same temperature?
\[ \tfrac{1}{2}\,\mathrm{NO}(\mathrm{g}) + \tfrac{1}{2}\,\mathrm{SO}_3(\mathrm{g}) \rightleftharpoons \tfrac{1}{2}\,\mathrm{NO}_2(\mathrm{g}) + \tfrac{1}{2}\,\mathrm{SO}_2(\mathrm{g}) \]
(A) \( \sqrt{650} \)
(B) \( \dfrac{1}{650} \)
(C) \( \dfrac{1}{\sqrt{650}} \)
(D) \( \dfrac{1}{2} \times 650 \)
▶️ Answer/Explanation
Detailed solution

The equilibrium constant depends on the way the chemical equation is written.

Original reaction: \[ \mathrm{NO}_2(\mathrm{g}) + \mathrm{SO}_2(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g}) + \mathrm{SO}_3(\mathrm{g}), \qquad K_{\mathrm{c}} = 650. \]

The new reaction is: \[ \tfrac{1}{2}\,\mathrm{NO}(\mathrm{g}) + \tfrac{1}{2}\,\mathrm{SO}_3(\mathrm{g}) \rightleftharpoons \tfrac{1}{2}\,\mathrm{NO}_2(\mathrm{g}) + \tfrac{1}{2}\,\mathrm{SO}_2(\mathrm{g}) \]

Two changes have been made to the original equation:

  • It has been reversed (products \(\leftrightarrow\) reactants), so \[ K_{\text{reversed}} = \dfrac{1}{K_{\mathrm{c}}} = \dfrac{1}{650}. \]
  • All coefficients have been multiplied by \(\tfrac{1}{2}\). When all coefficients are multiplied by a factor \(n\), \[ K_{\text{new}} = \left(K_{\text{old}}\right)^{n}. \] Here \(n = \tfrac{1}{2}\), so \[ K_{\text{final}} = \left( \dfrac{1}{650} \right)^{1/2} = \dfrac{1}{\sqrt{650}}. \]

✅ Answer: (C)

Question

Which equilibrium constant corresponds to the spontaneous reaction with the most negative value of \( \Delta G^{\theta} \)?
(A) \( 4.9 \times 10^{-3} \)
(B) \( 8.2 \times 10^{-3} \)
(C) \( 4.9 \times 10^{2} \)
(D) \( 8.2 \times 10^{2} \)
▶️ Answer/Explanation
Detailed solution

The relationship between standard Gibbs free energy change and the equilibrium constant is
\[ \Delta G^{\theta} = – R T \ln K \]

Here \( R \) is the gas constant, \( T \) is the absolute temperature, and \( K \) is the equilibrium constant.

For a fixed temperature \( T \), \( R T \) is constant, so \( \Delta G^{\theta} \) becomes more negative as \( \ln K \) becomes larger, that is, as \( K \) becomes larger.

Among the given values:

\( K_1 = 4.9 \times 10^{-3} \)
\( K_2 = 8.2 \times 10^{-3} \)
\( K_3 = 4.9 \times 10^{2} \)
\( K_4 = 8.2 \times 10^{2} \)

The largest equilibrium constant is \( 8.2 \times 10^{2} \), so this corresponds to the most negative \( \Delta G^{\theta} \) (most thermodynamically favorable reaction).

✅ Answer: (D)

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