Chemistry

1. 1.

   Explain the sign of the enthalpy change when a reaction warms its surroundings at constant pressure.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Define the system as the reacting chemicals. A warmer environment has received energy, so the system has lost enthalpy even though the thermometer reading rises.

   Marking points

   • Heat flows from the reacting system to the surroundings.
   • The reaction is exothermic.
   • The system's enthalpy change is negative.

   Examiner tip: The thermometer's temperature change and reaction enthalpy have opposite signs here.

2. 2.

   Why does a catalyst not change the enthalpy change of a reaction, even when it makes the reaction faster?

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Enthalpy change depends on the difference between final and initial states. Altering the barrier between those states affects rate, not that difference.

   Marking points

   • It provides an alternative pathway.
   • Activation energy is reduced.
   • Reactant and product enthalpy levels are unchanged.

   Examiner tip: Do not confuse activation energy with reaction enthalpy.

3. 3.

   A reaction heats 100 g of solution from 20.0 to 26.0 degrees C. Use c = 4.18 J g^-1 K^-1 and 0.0500 mol reacted. Neglect losses and calculate the molar reaction enthalpy.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Use q = mc deltaT = 100 x 4.18 x 6. The reaction supplies this heat, so negate it and divide by 0.0500 after converting joules to kilojoules.

   Marking points

   • Solution heat gain = 2508 J.
   • Reaction heat = -2.508 kJ.
   • Molar enthalpy = -50.2 kJ mol^-1 to three significant figures.

   Examiner tip: Use total solution mass, not the mass of just one reactant.

4. 4.

   Given deltaHf(CO2) = -394, deltaHf(H2O(l)) = -286 and deltaHf(CH4) = -75 kJ mol^-1, calculate deltaH for CH4 + 2O2 -> CO2 + 2H2O(l).

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The product total is -394 + 2(-286) = -966. Subtract the reactant total -75: -966 - (-75) = -891 kJ mol^-1.

   Marking points

   • O2 in its standard state has zero formation enthalpy.
   • Use sum(products) minus sum(reactants), including coefficients.
   • deltaH = -891 kJ mol^-1.

   Examiner tip: The water phase matters; do not substitute a gaseous-water value.

5. 5.

   For a reaction deltaH = +40.0 kJ mol^-1 and deltaS = +100 J mol^-1 K^-1, assumed constant. Calculate the temperature above which deltaG is negative, and explain why this does not guarantee a fast reaction.

   [4 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. At the threshold, set deltaG to zero and solve 40.0/0.100 = 400 K. A positive entropy change favours higher temperature, but feasibility describes the driving force rather than the pathway barrier.

   Marking points

   • Convert deltaS to 0.100 kJ mol^-1 K^-1.
   • Use deltaG = deltaH - T deltaS.
   • deltaG < 0 for T > 400 K.
   • A large activation barrier may still make the reaction slow.

   Examiner tip: Use kelvin, and distinguish zero Gibbs change at the threshold from negative above it.

6. 6.

   A combustion calorimeter gives a less negative enthalpy than a reliable reference. Explain how incomplete combustion and heat loss each cause this, and propose a distinct improvement for each.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Both errors reduce the magnitude of measured heat, but their causes differ. More oxygen addresses chemical conversion; insulation addresses where released energy goes.

   Marking points

   • Incomplete combustion releases less energy per mole of fuel than complete combustion.
   • Supply sufficient oxygen to improve completeness.
   • Heat lost to the room reduces measured temperature rise and calculated heat transfer.
   • Use insulation/a shield to reduce environmental heat loss, and a temperature-time cooling correction to account for residual loss.

   Examiner tip: Link each improvement to its own error mechanism.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.