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 

\(2\,\mathrm{mol}\) of \( \mathrm{NO(g)} \) and \(1\,\mathrm{mol}\) of \( \mathrm{Cl_2(g)} \) were mixed together, and the system was allowed to reach equilibrium at \(35^\circ\mathrm{C}\):

\( \mathrm{2NO(g)+Cl_2(g)\rightleftharpoons2NOCl(g)} \qquad K=6.5\times10^4 \text{ at }35^\circ\mathrm{C} \)

Which relationship is correct for this equilibrium at \(35^\circ\mathrm{C}\)?

(A) \( \mathrm{[NOCl]=[NO]} \)
(B) \( \mathrm{[NOCl]=2[Cl_2]} \)
(C) \( \mathrm{[NOCl]\gg[NO]} \)
(D) \( \mathrm{[NOCl]\ll[NO]} \)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{C}} \)

The equilibrium constant is very large:

\(K=6.5\times10^4\)

A large value of \(K\) indicates that equilibrium strongly favours the products.

For the reaction:

\( \mathrm{2NO+Cl_2\rightleftharpoons2NOCl} \)

the equilibrium mixture therefore contains a much greater concentration of \( \mathrm{NOCl} \) than \( \mathrm{NO} \).

Hence \( \mathrm{[NOCl]\gg[NO]} \), so the correct answer is (C).

Question 

Consider the equilibrium reaction:

\( \mathrm{CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)} \)

Under certain conditions, \(K=2.5\).

At the same temperature and pressure, a mixture has the following concentrations, in \( \mathrm{mol\,dm^{-3}} \):

\( \mathrm{[CO]=0.2\qquad[H_2]=0.4\qquad[CH_3OH]=0.32} \)

Which statement is correct?

(A) Mixture is at equilibrium and no reactions are occurring.
(B) Forward and reverse reactions are occurring at the same rate.
(C) Forward reaction rate is favoured to establish equilibrium.
(D) Reverse reaction is favoured to establish equilibrium.
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{D}} \)

Calculate the reaction quotient:

\( Q=\dfrac{\mathrm{[CH_3OH]}}{\mathrm{[CO][H_2]^2}} \)

\( Q=\dfrac{0.32}{(0.2)(0.4)^2} \)

\( Q=\dfrac{0.32}{0.032}=10 \)

Compare \(Q\) with \(K\):

\( Q=10>K=2.5 \)

There is too much product relative to the reactants, so the equilibrium shifts to the left. Therefore, the reverse reaction is favoured until \(Q\) decreases to \(K\).

Therefore, the correct answer is (D).

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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