Entropy Example Problem

This example problem demonstrates how to calculate the change in entropy of a system's surroundings following a chemical reaction at constant temperature and pressure.

Problem:

Calculate the entropy of the surroundings for the following two reactions.

a.) C₂H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4H₂O(g)
ΔH = -2045 kJ

b.) H₂O(l) → H₂O(g)
ΔH = +44 kJ

Solution

The change in entropy of the surroundings after a chemical reaction at constant pressure and temperature can be expressed by the formula

ΔS_surr = -ΔH/T

where
ΔS_surr is the change in entropy of the surroundings
-ΔH is heat of reaction
T = Absolute Temperature in Kelvin

Reaction a

ΔS_surr = -ΔH/T
ΔS_surr = -(-2045 kJ)/(25 + 273)
**Remember to convert °C to K**
ΔS_surr = 2045 kJ/298 K
ΔS_surr = 6.86 kJ/K or 6860 J/K

Note the increase in the surrounding entropy since the reaction was exothermic.

Reaction b

ΔS_surr = -ΔH/T
ΔS_surr = -(+44 kJ)/298 K
ΔS_surr = -0.15 kJ/K or -150 J/K

This reaction needed energy from the surroundings to proceed and reduced the entropy of the surroundings.

Answer:

The change in entropy of the surroundings of reaction 1 and 2 was 6860 J/K and -150 J/K respectively.