What is the standard entropy change (ΔS°) for the reaction: 2H2(g) + O2(g) → 2H2

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Question: What is the standard entropy change (ΔS°) for the reaction: 2H2(g) + O2(g) → 2H2O(g)?

Options:

Correct Answer: Negative

Solution:

The reaction results in a decrease in the number of gas molecules (from 3 to 2), leading to a negative change in entropy.

What is the standard entropy change (ΔS°) for the reaction: 2H2(g) + O2(g) → 2H2O(g)?

What is the standard entropy change (ΔS°) for the reaction: 2H2(g) + O2(g) → 2H2O(g)?

Step 1: Identify the reactants and products in the reaction: 2H2(g) + O2(g) → 2H2O(g).
Step 2: Count the number of gas molecules on the reactant side: There are 2 H2 and 1 O2, totaling 3 gas molecules.
Step 3: Count the number of gas molecules on the product side: There are 2 H2O molecules, totaling 2 gas molecules.
Step 4: Compare the number of gas molecules: The reactants have 3 gas molecules, and the products have 2 gas molecules.
Step 5: Determine the change in the number of gas molecules: Since the number of gas molecules decreases from 3 to 2, this indicates a decrease in disorder.
Step 6: Conclude that the standard entropy change (ΔS°) is negative due to the decrease in the number of gas molecules.

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