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# Which of the following reactions results in a decrease in entropy? (2 points) Question 17 options: 1) 6CO2 (g) + 6H2O (l) yields C6H12O6 (s) + 6O2 (g) 2) 2NH3 (g) yields N2 (g) + 3H2 (g) 3) N2O4 (g) yields 2NO2 (g) 4) 2H2O2 (aq) yields 2H2O (l) + O2 (g)

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Correct answer: Option a, that is, 6CO2 (g) + 6H2O (l)→C6H12O6 (s)+ 6O2 (g)Reason:the Entropy is the measure of the degree of disorder in a system. The greater the disorder, greater the entropy.The change in the entropy of the system can be predicted on the basis of a change in the number of moles of reactants and products. For example, if Δn = ∑n - ∑n reactants > o, an increase of the entropy.On the other hand, if Δn = ∑n - ∑n reactants < oh, it's going to decrease the entropy.Now consider the following casesOption 1: 6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g)Here, Δn = 7 - 12 = -5 . Therefore, the entropy of the system decreases.Option 2: 2NH3 (g) → N2 (g) + 3H2 (g)Here, Δn = 4 - 2 = 2 . Therefore, the entropy of the system increases.Option 3: N2O4 (g) → 2NO2 (g)Here, Δn = 2 - 1 = 1 . Therefore, the entropy of the system increases.option 4: 2H2O2 (aq) → 2H2O (l) + O2 (g)Here, Δn = 3 - 2 = 1 . Therefore, the entropy of the system increases.