Home › Subjects › A-Level H2 Chemistry › Enthalpy, entropy and the feasibility of reactions
A-Level Enthalpy, entropy and the feasibility of reactions
What the A-Level syllabus expects for Enthalpy, entropy and the feasibility of reactions, and how to practise it.
What the syllabus expects
- Explain that nearly every reaction exchanges energy, chiefly as heat tied to bonds breaking and forming, and can be exothermic (delta H negative) or endothermic (delta H positive).
- Draw and read an energy profile diagram, relating it to the reaction's enthalpy change and its activation energy.
Scope: see also Section 8 - Explain and make use of these terms:
- enthalpy change of reaction under standard conditions, covering formation, combustion, hydration, solution, neutralisation and atomisation
- bond energy, positive because a bond costs energy to break
- lattice energy, negative as gaseous ions gather into a solid lattice
- Compute enthalpy changes from suitable experimental data, drawing on heat change = mc delta T.
- Account qualitatively for how an ion's charge and its radius affect how large the lattice energy turns out.
- Use Hess' Law to build simple energy cycles such as the Born-Haber cycle, and calculate with them using the relevant terms, ionisation energy and electron affinity included, focusing on:
- reaching enthalpy changes that direct experiment cannot, for example a formation enthalpy found from combustion enthalpies
- how a simple ionic solid comes to form, and how its aqueous solution does
- average bond energies
- Explain what entropy is and put the term to use.
- Discuss how each of the following alters a chemical system's entropy:
Scope: no quantitative treatment is expected - a shift in temperature
- a change of phase
- a change in the number of particles, gaseous systems especially
- Forecast whether a given process or reaction raises or lowers entropy.
- Quote and apply the standard Gibbs free energy relationship, delta G-standard = delta H-standard minus T times delta S-standard.
Scope: you need not calculate delta S-standard from standard entropies - Judge from the sign of delta G-standard whether a reaction or process runs spontaneously.
- Appreciate where delta G-standard falls short as a guide to whether a reaction proceeds.
- Given standard enthalpy and entropy changes, predict how altering the temperature bears on spontaneity.
How it's examined
Questions on this topic most often ask you to calculate, state, deduce, define. About 8% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (3 marks)
Using Data Booklet values plus the figures below, find the mean Al–Cl bond energy in Al2Cl6. The data are: a dimerisation enthalpy for Al2Cl6(g) of −158.8 kJ mol−1, a formation enthalpy for AlCl3(g) of −248.6 kJ mol−1, and an atomisation enthalpy for Al(s) of +326 kJ mol−1. Draw an energy cycle and calculate the average Al–Cl bond energy.
Show the worked answer
Build an energy cycle linking Al2Cl6(g) and its gaseous atoms via the elements. Formation of Al2Cl6(g) from elements = 2 x (formation of AlCl3(g)) + dimerisation = 2(-248.6) + (-158.8) = -656.0 kJ mol⁻¹. Atomising the same elements to gaseous atoms (2Al(s) + 3Cl2(g) -> 2Al(g) + 6Cl(g)): 2Al(s)->2Al(g) = 2(+326) = +652 ; 3Cl2->6Cl (Cl-Cl = +244) = 3(+244) = +732 ; total = +1384 kJ. By Hess' Law, atomising Al2Cl6(g) -> 2Al(g) + 6Cl(g): ΔH = -(-656.0) + 1384 = +2040 kJ mol⁻¹ (energy to break all Al-Cl bonds). Al2Cl6 is a chlorine-bridged dimer containing 8 Al-Cl bonds, so mean Al-Cl bond energy = 2040 / 8 = 255 kJ mol⁻¹.
Example 2 (2 marks)
Using bond energy values from the Data Booklet, work out C2H2's standard enthalpy change of combustion.
Show the worked answer
C2H2 + 5/2 O2 -> 2CO2 + H2O Bonds broken (energy absorbed): 1 C(triple)C = 840 2 C-H = 2 x 410 = 820 2.5 O=O = 2.5 x 496 = 1240 total = 2900 kJ Bonds formed (energy released): 4 C=O (in CO2) = 4 x 805 = 3220 2 O-H = 2 x 460 = 920 total = 4140 kJ DeltaH_c = bonds broken - bonds formed = 2900 - 4140 = -1240 kJ/mol
Example 3 (2 marks)
Given that TiCl3 decomposes spontaneously only at high temperature, deduce the sign of the enthalpy change and explain your reasoning.
Show the worked answer
For a reaction to be spontaneous, delta G = delta H - T delta S must be negative. Decomposition of a solid TiCl3 into products increases disorder (typically producing gas), so delta S is positive. If the reaction is spontaneous only at high temperature, then at low temperature it is non-spontaneous (delta G > 0) and only at high T does the -T delta S term become large enough and negative to make delta G < 0. This behaviour (non-spontaneous at low T, spontaneous at high T) requires delta H to be positive: with delta H > 0 and delta S > 0, delta G = delta H - T delta S is positive at low T and becomes negative only when T is large. Therefore delta H is positive (endothermic).
More worked questions on this topic
- Take the equilibrium: N2 (g) + 3H2 (g) ⇌ 2NH3 (g). Suppose the ammonia formed were a liquid rat (2 marks)
- Calcium carbide reacts violently, even explosively, with water. Take the standard enthalpies of (2 marks)
- In a further neutralisation, HCl at the same concentration as that used in (a)(iii) was fully n (2 marks)
- When HCl is neutralised by NaOH the NaCl produced has a standard enthalpy of formation of −411 (2 marks)
- For the equilibrium N2(g) + 3H2(g) ⇌ 2NH3(g), work out the temperature (in K) above or below wh (2 marks)
More A-Level H2 Chemistry topics
The make-up of the atom and how its electrons are arranged · How atoms bond and how that governs a substance's behaviour · Ideal gases and working with gas mixtures · Competing definitions of acids and bases · Trends in the elements across a period and down a group · The mole and reacting-quantity calculations · all of A-Level H2 Chemistry