Drivers of Reactions, complete assessment covering enthalpy, calorimetry, bond energy, Hess Law, entropy and Gibbs free energy from L01-L13. 15 MC questions (auto-marked) + 5 written questions (self-marked). Complete all questions before submitting.
For an exothermic reaction, the enthalpy change is:
On an energy profile diagram, activation energy is the energy difference between:
In q = mcΔT for a water calorimetry experiment, m usually represents:
The enthalpy of neutralisation for a strong acid and strong base is similar because:
Dissolving an ionic solid is endothermic overall when:
A catalyst changes the energy profile by:
Using average bond energies, ΔH is estimated by:
Standard enthalpy of formation refers to forming:
Hess Law works because enthalpy is:
Photosynthesis is energetically opposite to respiration because photosynthesis:
Heat of combustion is usually reported as:
A reaction with gases produced from aqueous or solid reactants often has positive ΔS because:
Standard entropy change for a reaction is calculated using:
A reaction is spontaneous at constant temperature and pressure when:
For ΔG = ΔH - TΔS, increasing temperature favours spontaneity most directly when:
A 100.0 g sample of water increases from 21.0 C to 29.4 C when a fuel burns. Calculate the heat absorbed by the water and explain how this relates to the enthalpy change of the fuel. Use c = 4.18 J g-1 C-1.
ΔT = 29.4 - 21.0 = 8.4 C. q = mcΔT = 100.0 x 4.18 x 8.4 = 3511 J = 3.51 kJ absorbed by the water. The fuel released approximately this amount of heat to the surroundings, so the fuel combustion enthalpy for the measured sample is negative. In practice, heat loss to the environment means the experimental value is usually less exothermic than the true value.
Marks: 1, temperature change | 1, q substitution | 1, correct heat value | 1, sign linked to exothermic fuel | 1, heat loss limitationExplain why a catalyst increases rate but does not change ΔH. Then outline how bond energies can estimate ΔH for a reaction.
A catalyst provides an alternative reaction pathway with lower activation energy, so more collisions are successful. It does not change the energies of the reactants or products, so ΔH is unchanged. Bond energy estimates compare energy absorbed to break bonds in reactants with energy released when product bonds form: ΔH = sum of bonds broken - sum of bonds formed. If stronger bonds are formed than broken, the reaction is exothermic.
Marks: 1, catalyst lowers Ea | 1, alternate pathway | 1, reactant/product energies unchanged | 1, bond formula | 1, exothermic interpretationUse Hess Law to explain how a target enthalpy change can be found from known equations, including what happens when equations are reversed or multiplied.
Hess Law states that enthalpy change depends only on initial and final states, so reaction steps can be added to obtain a target reaction. If a known equation is reversed, the sign of ΔH is reversed. If an equation is multiplied, its ΔH is multiplied by the same factor. The manipulated equations are added, cancelling species that appear on both sides, and the adjusted ΔH values are summed to give the target enthalpy change.
Marks: 1, state function idea | 1, reverse sign rule | 1, multiply scaling rule | 1, adding/cancelling equations | 1, summing adjusted enthalpiesA reaction has ΔH = +45 kJ mol-1 and ΔS = +160 J K-1 mol-1. Calculate ΔG at 298 K and predict whether it is spontaneous at 298 K.
Convert entropy to kJ: ΔS = 160 J K-1 mol-1 = 0.160 kJ K-1 mol-1. ΔG = ΔH - TΔS = 45 - (298 x 0.160) = 45 - 47.68 = -2.68 kJ mol-1. Since ΔG is negative, the reaction is spontaneous at 298 K under the stated conditions, even though it is endothermic, because the positive entropy term is large enough.
Marks: 1, entropy conversion | 1, correct substitution | 1, calculated ΔG | 1, spontaneous prediction | 1, explanation of entropy offsetting endothermic ΔHCompare enthalpy and entropy as drivers of chemical change, and explain why ΔG gives a fuller prediction than either alone.
Enthalpy describes heat absorbed or released by a reaction, so exothermic reactions are often favoured because they lower chemical potential energy. Entropy describes energy and particle dispersal, so reactions that increase disorder or accessible arrangements are also favoured. Neither factor alone is sufficient, because an endothermic reaction can be spontaneous if entropy gain is large, and an exothermic reaction can be non-spontaneous if entropy loss is large. Gibbs free energy combines both terms through ΔG = ΔH - TΔS and includes temperature, giving the spontaneity criterion ΔG < 0.
Marks: 1, enthalpy described | 1, entropy described | 1, limitation of enthalpy alone | 1, limitation of entropy alone | 1, Gibbs criterion with temperature