Science, Tech, Math › Science Calculate the Change in Entropy From Heat of Reaction Entropy Example Problem Share Flipboard Email Print PM Images / Getty Images Science Chemistry Basics Chemical Laws Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry In Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate By Anne Marie Helmenstine, Ph.D. Chemistry Expert Ph.D., Biomedical Sciences, University of Tennessee at Knoxville B.A., Physics and Mathematics, Hastings College Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. She has taught science courses at the high school, college, and graduate levels. our editorial process Facebook Facebook Twitter Twitter Anne Marie Helmenstine, Ph.D. Updated December 28, 2018 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 kJb.) H2O(l) → H2O(g)ΔH = +44 kJSolutionThe change in entropy of the surroundings after a chemical reaction at constant pressure and temperature can be expressed by the formulaΔSsurr = -ΔH/TwhereΔSsurr is the change in entropy of the surroundings-ΔH is heat of reactionT = Absolute Temperature in KelvinReaction aΔSsurr = -ΔH/TΔSsurr = -(-2045 kJ)/(25 + 273)**Remember to convert °C to K**ΔSsurr = 2045 kJ/298 KΔSsurr = 6.86 kJ/K or 6860 J/KNote 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 bΔSsurr = -ΔH/TΔSsurr = -(+44 kJ)/298 KΔSsurr = -0.15 kJ/K or -150 J/KThis 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.Answer:The change in entropy of the surroundings of reaction 1 and 2 was 6860 J/K and -150 J/K respectively.