Exercise practice: Thermodynamics

5.1 How the exercises are grouped

The chapter prints exercises 5.1 to 5.22. The exercise practice page groups them by skill and works a representative question in each group.

5.2 Group A: the first law and work

Exercise 5.1 identifies the state function. A quantity independent of the path is a state function, so internal energy and enthalpy qualify; heat and work do not.

Exercise 5.2 gives the adiabatic condition. An adiabatic process exchanges no heat with the surroundings, so q = 0. Exercise 5.7 gives q = +701 J and a work term of -394 J, because the system does 394 J of work on the surroundings. Adding the two gives an internal energy change of +307 J.

Exercise 5.9 uses heat capacity. Heating 60.0 g of aluminium from 35 to 55 degrees Celsius means a temperature rise of 20 K. The amount of aluminium is 60.0/27 = 2.22 mol, and with a molar heat capacity of 24 J mol^-1 K^-1 the heat is 2.22 x 24 x 20 = 1.07 x 10^3 J, about 1.1 kJ.

Exercise 5.16 asks about an isolated system with delta U = 0. For an isolated system no heat and no work cross the boundary, so nothing changes and delta S is zero for a reversible process; the exercise is designed to show that delta U = 0 does not by itself fix delta S, because a free expansion into vacuum also has delta U = 0 for an ideal gas while its entropy increases.

5.3 Group B: thermochemistry

Exercise 5.3 relies on the convention that an element in its standard state carries a formation enthalpy of zero. Exercise 5.4 compares delta U and delta H for the combustion of methane. The combustion consumes gas, so delta n_g is negative and delta H is less than delta U.

Exercise 5.5 combines combustion enthalpies to find the enthalpy of formation of methane. Write the formation equation, then use Hess's law with the given combustion equations for methane, graphite and hydrogen; the answer is -74.8 kJ mol^-1.

Exercise 5.8 uses a bomb calorimeter, so the measured quantity is delta U. The reaction NH2CN + 3/2 O2 giving N2 + CO2 + H2O changes the gas moles by 2 minus 1.5, which is +0.5, so delta H is delta U plus 0.5 RT. With delta U = -742.7 kJ mol^-1 at 298 K the enthalpy change is about -741.5 kJ mol^-1.

Exercise 5.10 asks for the enthalpy change on freezing one mole of water at 273 K. Freezing is the reverse of fusion, so the enthalpy change is the negative of the enthalpy of fusion, and the entropy change is the negative of the fusion entropy.

Exercise 5.13 uses the formation of ammonia. The reaction N2 + 3H2 giving 2NH3 has a standard enthalpy change of -92.4 kJ mol^-1, so the standard enthalpy of formation of one mole of ammonia is half of that, -46.2 kJ mol^-1.

Exercise 5.14 builds the formation enthalpy of methanol from its combustion and the formation enthalpies of carbon dioxide and water. Writing the formation equation and subtracting the combustion data gives about -239 kJ mol^-1.

Exercise 5.15 finds the C-Cl bond enthalpy in carbon tetrachloride. The process CCl4(g) giving C(g) + 4Cl(g) requires the enthalpy of atomisation of carbon plus twice the bond enthalpy of chlorine minus the formation enthalpy of CCl4, after correcting for vaporisation. Dividing the total by four gives the bond enthalpy per C-Cl bond.

5.4 Group C: entropy and the second law

Exercise 5.6 combines a reaction with a positive entropy change and a positive enthalpy change. The reaction is possible at high temperature, where the T delta S term outweighs the enthalpy term and delta G becomes negative.

Exercise 5.18 asks for the signs of delta H and delta S for the combination of two chlorine atoms into a molecule. A bond forms, so enthalpy is released and delta H is negative. Two gas particles become one, so the disorder falls and delta S is negative.

Exercise 5.22 asks for the entropy change of the surroundings when one mole of liquid water forms from its elements under standard conditions. The formation enthalpy is -286 kJ mol^-1, so the surroundings gain that amount of heat and their entropy change is +286000/298, which is about +960 J K^-1 mol^-1.

5.5 Group D: Gibbs energy

Exercise 5.17 finds the temperature above which a reaction becomes spontaneous. The reaction has delta H = 400 kJ mol^-1 and delta S = 0.2 kJ K^-1 mol^-1, so delta G becomes negative above T = 400/0.2 = 2000 K.

Exercise 5.19 combines delta U, the gas mole change and delta S to find delta G. First convert delta U to delta H, then use delta G = delta H - T delta S, and decide spontaneity from the sign.

Exercise 5.20 uses delta G standard equal to minus RT ln K. With K = 10 at 300 K, delta G standard is -8.314 x 300 x ln 10, which is about -5.7 kJ mol^-1.

Exercise 5.21 discusses the thermodynamic stability of nitric oxide. The formation of NO from its elements is endothermic, so NO is thermodynamically unstable with respect to decomposition into nitrogen and oxygen, but the decomposition is very slow because the activation energy is high. The reaction of NO with oxygen to form NO2 is exothermic, which explains why NO is readily oxidised in air.