Edexcel A Level Chemistry:复习笔记5.4.2 Entropy Calculations

Total Entropy Calculations


  • If we take the reaction between sodium and chorine, we know that this is a very exothermic reaction and also involves a decrease in entropy as a solid is produced from a solid and a gas (provided a flame is used to supply the necessary activation energy)

2Na (s) + Cl2 (g) → 2NaCl (s)

  • If there is a decrease in entropy, how can the reaction be spontaneous?
  • We need to take into account the entropy of the surroundings as well
    • The energy being released causes a substantial increase in entropy of the surroundings because there are more ways of arranging the quanta (packets of energy) in the surroundings than the system alone
  • Therefore, the total entropy change for a reaction is

ΔSΘ total = ΔS Θsys + ΔSΘsurr

(sys = system and surr = surroundings)

  • So, in the case of sodium and chlorine, the large amounts of energy released makes ΔSΘsurr very positive, which will outweigh the negative value of ΔS Θsys

Worked Example

Calculating total entropy change

Calculate the total entropy change in the formation of 1 mole of sodium chloride from its elements in their standard state

ΔSΘsys = -90.1 J K-1 mol-1

ΔSΘsurr = +1379 J K-1 mol-1


ΔS Θtotal = ΔS Θsys + ΔS Θsurr

ΔSΘ total = -90.1 + 1379 = 1289 J K-1 mol-1

Entropy Change in the System

  • Entropy changes are an order of magnitude smaller than enthalpy changes, so entropy is measured in joules rather than kilojoules. The full unit for entropy is J K-1 mol-1
  • The standard entropy change (ΔSΘsystem) for a given reaction can be calculated using the standard entropies (Sꝋ ) of the reactants and products
  • The equation to calculate the standard entropy change of a system is:

ΔSΘsystem = ΣΔSΘproducts - ΣΔSΘreactants

(where Σ = sum of)

  • For example, the standard entropy change for the formation of ammonia (NH3) from nitrogen (N2) and hydrogen (H2) can be calculated using this equation

N2(g) + 3H2(g) ⇋ 2NH3(g)

ΔSΘsystem = (2 x ΔSΘ(NH3)) - (ΔSΘ(N2) + 3 x ΔSΘ(H2))

  • Notice that, unlike enthalpy of formation for elements, entropy for elements is not zero and you can find entropy values for elements and compounds in data books

Worked Example

Calculating entropy changes

Calculate the entropy change of the system for the following reaction:

2Mg (s) + O2 (g) → 2MgO (s)

SΘꝋ[Mg(s)] = 32.60 J K-1 mol-1

SΘꝋ[O2(g)] = 205.0 J K-1 mol-1

SΘꝋ[MgO(s)] = 38.20 J K-1 mol-1


ΔSΘsystem = ΣΔSΘproducts - ΣΔSΘreactants

ΔSΘsystem = (2 x 38.20) - (2 x 32.60 + 205.0)

= -193.8 J K-1 mol-1

Worked Example

Calculating entropy changes

What is the entropy change when ammonia is formed from nitrogen and hydrogen?

N2 (g) + 3H2 (g) ⇌ 2NH3 (g)

SΘꝋ[N2 (g)] = 191.6 J K-1 mol–1

SΘꝋ[H2 (g)] = 131 J K-1 mol–1

SΘꝋ[NH3] = 192.3 J K-1 mol–1


ΔSΘsystem = ΣΔSΘproducts - ΣΔSΘreactants

ΔSΘsystem= [2 x SΘ(NH3)] - [SΘ(N2)+ (3 x SΘ(H2 ))]

ΔSΘsystem= [2 x 192.3] - [191.6 + (3 x 131)]

ΔSΘsystem = 384.6 - 584.6

ΔSΘsystem= -200 J K-1 mol–1

Exam Tip

Use the stoichiometry  of the equation and the correct state of the compounds when calculating the entropy change of a reaction.

Entropy Change in the Surroundings

  • To calculate the entropy change of the surroundings, ΔSΘ surr , we need to know the energy that has been transferred to them
  • This is given by the enthalpy change, ΔH, and the relationship can be expressed as:

  • T is the absolute temperature
  • The entropy change of the surroundings depends upon temperature
    • The transfer of a given quantity of energy to surroundings at a low temperature will produce a greater entropy change than the transfer of the same amount of energy to the surroundings at a higher temperature

Worked Example

Calculating entropy of surroundings

Calculating entropy of surroundings for the reaction between aluminium oxide and carbon at 298 K

Al2O3 (s) + 3C (s) → 2Al (s) + 3CO (g)

ΔHΘ = +1336 kJ mol-1

T = 298 K


    • Convert ΔH from kJ mol-1 to J mol-1 by multiplying by 1000
    • = - 4483 J K-1 mol-1