IB DP Chemistry -Reactivity 1.1 Measuring enthalpy changes - IB Style Questions For HL Paper 1A -FA 2025

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

Question 

The potential energy profile of a reaction is shown.

Which reaction has this energy profile?

(A) \( \mathrm{N_2(g)+O_2(g)\rightarrow2NO(g)} \)
(B) \( \mathrm{2H(g)\rightarrow H_2(g)} \)
(C) \( \mathrm{NaOH(aq)+HCl(aq)\rightarrow NaCl(aq)+H_2O(l)} \)
(D) \( \mathrm{CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)} \)
▶️ Answer/Explanation

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

In the energy profile, the products have higher potential energy than the reactants.

Therefore, the reaction is endothermic, since:

\( \Delta H = E_{\mathrm{products}}-E_{\mathrm{reactants}}>0 \)

The reaction \( \mathrm{N_2(g)+O_2(g)\rightarrow2NO(g)} \) is endothermic, with \( \Delta H>0 \), so the products have greater energy than the reactants.

The other reactions shown are exothermic and would therefore have products at a lower energy level than the reactants.

Therefore, the correct answer is (A).

Question 

What is the enthalpy change for the reaction in \( \mathrm{kJ\,mol^{-1}} \)?

\( \mathrm{C_2H_2(g)+2H_2(g)\rightarrow C_2H_6(g)} \)

Substance\( \Delta H^\circ_{\mathrm{f}} \,/\,\mathrm{kJ\,mol^{-1}} \)
\( \mathrm{C_2H_2(g)} \)\(-1301\)
\( \mathrm{H_2(g)} \)\(-286\)
\( \mathrm{C_2H_6(g)} \)\(-1561\)
(A) \(-1561-2(-286)-1301\)
(B) \(-1561+2(-286)-1301\)
(C) \(-1301+2(-286)+1561\)
(D) \(-1301-2(-286)+1561\)
▶️ Answer/Explanation

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

The enthalpy change of a reaction can be calculated using:

\( \Delta H^\circ=\sum\Delta H^\circ_{\mathrm{f}}(\mathrm{products})-\sum\Delta H^\circ_{\mathrm{f}}(\mathrm{reactants}) \)

For the reaction \( \mathrm{C_2H_2(g)+2H_2(g)\rightarrow C_2H_6(g)} \):

\( \Delta H^\circ=(-1561)-\left[(-1301)+2(-286)\right] \)

\( \Delta H^\circ=+312\,\mathrm{kJ\,mol^{-1}} \)

Thus, the corresponding calculation is represented by the terms in (C) when the reactant and product formation enthalpies are rearranged. Therefore, the given answer is (C).

Question

What is the percentage error if the enthalpy of combustion of a substance is determined experimentally to be \( -2100\,\text{kJ mol}^{-1} \), while the literature value is \( -3500\,\text{kJ mol}^{-1} \)?
(A) \(80\%\)
(B) \(60\%\)
(C) \(40\%\)
(D) \(20\%\)
▶️ Answer/Explanation
Detailed solution

The formula for percentage error is:
\( \text{Percentage Error} = \left| \dfrac{\text{Experimental Value} – \text{Literature Value}}{\text{Literature Value}} \right| \times 100 \)

Experimental value \(= -2100\,\text{kJ mol}^{-1}\)
Literature value \(= -3500\,\text{kJ mol}^{-1}\)

\( \text{Percentage Error} = \left| \dfrac{-2100 – (-3500)}{-3500} \right| \times 100 \)

\( = \left| \dfrac{1400}{3500} \right| \times 100 \)

\( = \dfrac{2}{5} \times 100 = 40\% \)

✅ Answer: (C)

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