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Ch.18 - Free Energy and Thermodynamics
Tro - Chemistry: A Molecular Approach 4th Edition
Tro4th EditionChemistry: A Molecular ApproachISBN: 9780134112831Non è quello che usi tu?Cambia libro di testo
Capitolo 18, Problema 99a

Indicate and explain the sign of ΔSuniv for each process. a. 2 H2(g) + O2(g) → 2 H2O (l) at 298 K.

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Step 1: Understand the concept of entropy (ΔS). Entropy is a measure of the randomness or disorder of a system. The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time, and is constant if and only if all processes are reversible. Isolated systems spontaneously evolve towards thermodynamic equilibrium, the state with maximum entropy.
Step 2: Identify the state of the reactants and products. In this reaction, gaseous reactants (2 H2 and O2) are converted into a liquid product (2 H2O).
Step 3: Consider the change in entropy of the system (ΔSsys). The conversion of gas molecules to liquid molecules results in a decrease in randomness, as molecules in a liquid are more ordered than in a gas. Therefore, the entropy of the system decreases (ΔSsys is negative).
Step 4: Consider the change in entropy of the surroundings (ΔSsurr). The reaction is exothermic, meaning it releases heat into the surroundings. This increases the randomness of the surroundings, so the entropy of the surroundings increases (ΔSsurr is positive).
Step 5: Calculate the total change in entropy (ΔSuniv = ΔSsys + ΔSsurr). If the increase in entropy of the surroundings is greater than the decrease in entropy of the system, the total entropy of the universe increases (ΔSuniv is positive), and the process is spontaneous. If not, the total entropy of the universe decreases (ΔSuniv is negative), and the process is non-spontaneous. In this case, you would need to calculate the specific values of ΔSsys and ΔSsurr to determine the sign of ΔSuniv.

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