Calculate the Change in Entropy From Heat of Reaction

Entropy Example Problem

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The term "entropy" refers to disorder or chaos in a system. The greater the entropy, the greater the disorder. Entropy exists in physics and chemistry, but can also be said to exist in human organizations or situations. In general, systems tend toward greater entropy; in fact, according to the second law of thermodynamics, the entropy of an isolated system can never spontaneously decrease. 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.

What Change in Entropy Means

First, notice you never calculate entropy, S, but rather change in entropy, ΔS. This is a measure of the disorder or randomness in a system. When ΔS is positive it means the surroundings increased entropy. The reaction was exothermic or exergonic (assuming energy can be released in forms besides heat). When heat is released, the energy increases the motion of atoms and molecules, leading to increased disorder.

When ΔS is negative it means entropy of the surroundings were reduced or that the surroundings gained order. A negative change in entropy draws heat (endothermic) or energy (endergonic) from the surroundings, which reduces the randomness or chaos.

An important point to keep in mind is that the values for ΔS are for the surroundings! It's a matter of point of view. If you change liquid water into water vapor, entropy increases for the water, even though it decreases for the surroundings. It's even more confusing if you consider a combustion reaction. On the one hand, it seems breaking a fuel into its components would increase disorder, yet the reaction also includes oxygen, which forms other molecules.

Entropy Example

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
The change in entropy of the surroundings after a chemical reaction at constant pressure and temperature can be expressed by the formula
ΔSsurr = -ΔH/T
ΔSsurr is the change in entropy of the surroundings
-ΔH is heat of reaction
T = Absolute Temperature in Kelvin
Reaction a
ΔSsurr = -ΔH/T
ΔSsurr = -(-2045 kJ)/(25 + 273)
**Remember to convert °C to K**
ΔSsurr = 2045 kJ/298 K
ΔSsurr = 6.86 kJ/K or 6860 J/K
Note the increase in the surrounding entropy since the reaction was exothermic. An exothermic reaction is indicated by a positive ΔS value. This means heat was released to the surroundings or that the environment gained energy. This reaction is an example of a combustion reaction. If you recognize this reaction type, you should always expect an exothermic reaction and positive change in entropy.
Reaction b
ΔSsurr = -ΔH/T
ΔSsurr = -(+44 kJ)/298 K
ΔSsurr = -0.15 kJ/K or -150 J/K
This reaction needed energy from the surroundings to proceed and reduced the entropy of the surroundings. A negative ΔS value indicates an endothermic reaction occurred, which absorbed heat from the surroundings.
The change in entropy of the surroundings of reaction 1 and 2 was 6860 J/K and -150 J/K respectively.

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Your Citation
Helmenstine, Anne Marie, Ph.D. "Calculate the Change in Entropy From Heat of Reaction." ThoughtCo, Aug. 25, 2020, Helmenstine, Anne Marie, Ph.D. (2020, August 25). Calculate the Change in Entropy From Heat of Reaction. Retrieved from Helmenstine, Anne Marie, Ph.D. "Calculate the Change in Entropy From Heat of Reaction." ThoughtCo. (accessed February 9, 2023).