Entropy Calculations (Edexcel A Level Chemistry)

• 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

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

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

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

• 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

Answer

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

 

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