5.4.2 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

ΔS^Θ _total = ΔS ^Θ_sys + ΔS^Θ_surr

(sys = system and surr = surroundings) 

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

Worked Example

Answer

ΔS ^Θ_total = ΔS ^Θ_sys + ΔS ^Θ_surr

ΔS^Θ _total = -90.1 + 1379 = 1289 J K^-1 mol^-1

• 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 (ΔS^Θ_system) for a given reaction can be calculated using the standard entropies (S^ꝋ ) of the reactants and products
• The equation to calculate the standard entropy change of a system is:

ΔS^Θ_system = ΣΔS^Θ_products - ΣΔS^Θ_reactants

(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)

ΔS^Θ_system = (2 x ΔS^Θ(NH₃)) - (ΔS^Θ(N₂) + 3 x ΔS^Θ(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

Answer

ΔS^Θ_system = ΣΔS^Θ_products - ΣΔS^Θ_reactants

ΔS^Θ_system = (2 x 38.20) - (2 x 32.60 + 205.0)

= -193.8 J K^-1 mol^-1

Worked Example

Answer:

ΔS^Θ_system = ΣΔS^Θ_products - ΣΔS^Θ_reactants

ΔS^Θ_system= [2 x S^Θ(NH₃)] - [S^Θ(N₂)+ (3 x S^Θ(H₂ ))]

ΔS^Θ_system= [2 x 192.3] - [191.6 + (3 x 131)]

ΔS^Θ_system = 384.6 - 584.6

ΔS^Θ_system= -200 J K^-1 mol^–1

Exam Tip

• To calculate the entropy change of the surroundings, ΔS^Θ _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

Answer

• • 
  • Convert ΔH from kJ mol^-1 to J mol^-1 by multiplying by 1000 

 

= - 4483 J K^-1 mol^-1