Example 5.1. Calculating heat of a reaction

Problem: The thermochemical equation for the reaction of zinc with sulfur is

Zn(s) + S(s) → ZnS(s)

ΔHr = – 250.4 kJ

Calculate the heat, liberated when 8.0 g of sulfur react according to this equation.

Solution: The H_r given is the energy liberated when 1 mol S reacts. The molar mass of S is 32.0 g/mol, so

32.0 g S  – 250.4 kJ

8.0 g S  x kJ

x = 8.0(– 250.4)/32.0 = – 63 kJ

Answer: in reaction of 8.0 g S with zinc 63 kJ are liberated.

Standard conditions. Most thermochemical equations are written for the standard conditions (298 K and 1 atm) applied both to reagents and products. Heat effect of a reaction under the standard conditions is denoted as H_r°. If the temperature and pressure are different from 298 K and 1 atm they should be specified in the thermochemical equation.

Standard state of a substance is its only or the most stable state at standard conditions. Under these conditions a substance in its most stable aggregate state (liquid for H₂O) and most stable allotropic form (O₂ for oxygen) or crystalline polymorph (graphite for C) is said to be in a standard state.

The standard enthalpy of formation of a substance, H_f°, is defined as the heat associated with the formation of one mole of the compound from elementary substances in their standard states. For example, the standard enthalpy of liquid water formation is the heat of reaction:

H₂(g) + ½ O₂(g) → H₂O(l); H °_reaction = H_f ° H₂O(l) = –286 kJ

In accordance with the definition of the H_f°, enthalpies of the elementary substances in their standard states are equal to zero. The standard enthalpy of formation is a fundamental property of any compound. Most chemistry textbooks contain tables of H_f ° values.

Calculating of heat effects of reactions. Laws of thermochemistry are based on the fundamental law of conservation of matter⁴. Heat effects of reactions including supposed or even impossible ones can be calculated using the laws of thermochemistry.

The first law of thermochemistry states that heat effect of a reaction is equal in magnitude but opposite in sign to ΔH for the reverse reaction. For example:

Al(s) + 3/2 Br2(l) → AlBr3(s)	ΔHr = – 511 kJ
AlBr3(s) → Al(s) + 3/2 Br2(l)	ΔHr = 511 kJ

The second law of thermochemistry also called Hess's law states that ΔH_r is the same whether a reaction occurs in one step or in a series of steps. For example, excess of phosphorous react with chlorine to form PCl₃. Then PCl₃ can react with excess of chlorine to form PCl₅. The phosphorous pentachloride can be also synthesized in one step

	P Cl3
ΔH1		ΔH2
	ΔH3
P		PCl5

P(s) + 3/2 Cl2(g) → PCl3(g)	ΔH1 = – 280 kJ
PCl3(g) + Cl2(g) → PCl5(g)	ΔH2 = – 87 kJ
P(s) + 5/2 Cl2(g) → PCl5(g)	ΔH3 = – 367 kJ

As ΔH_r must be independent of the path of a reaction,

ΔH₃ = ΔH₁ + ΔH₂ = ΔH₁ + ΔH₂ = – 280 + (– 87) = – 367 kJ

  Hess's law can be applied to calculate the enthalpy change that can not be measured directly. For example, both graphite and diamond (the two crystalline forms of carbon) can be oxidized by an excess of oxygen to carbon dioxide.

C_(graphite) + O_2(g) → CO_2(g)     ΔH° = –393.51 kJ

C_(diamond) + O_2(g) → CO₂    ΔH° = –395.40 kJ

The standard enthalpy changes for these reactions can be combined to obtain ΔH° for the phase transformation between graphite and diamond. Subtraction of the second reaction from the first (i.e., writing the second equation in reverse and adding it to the first one) yields

C_(graphite) → C_(diamond)     ΔH° = 1.89 kJ

Hess’ law allows predicting the change in the enthalpy of huge numbers of chemical reactions from a relatively small base of experimental data.