In which of the following processes is the end product more ordered than the reactants? (in which process does entropy decrease.) select one: a. a. 2h2 (gas) + o2 (gas) = 2 h2o (gas) b. b. 2c8h18 (liquid) + 25o2 (gas) = 16co2 (gas) + 18h2o(gas) c. c. c6h12o6 (solid) + 6o2 (gas) = 6co2 (gas) + 6h2o (liquid) d. d. 2c7h5n3o6 (solid) = 3n2(gas) + 7co(gas) + 5h2o(gas) + 7 c(solid) e. e. nahco3(solid) + c2h4o2(liquid)=c2h3nao2(solid)+ h2o(liquid) + co2(gas)

Answer is A.

Explanation:

The change in entropy in a chemical reaction is lower when the moles of the gas product decrease. In the same way, the change in entropy is higher when the moles of the gas product increase.

For the reactions:

A. 2H₂(gas) + O₂(gas) → 2H₂O(gas). The entropy decrease because there are 4 moles of gas reactants and 2 moles of gas products.

B. 2C₈H₁₈(liquid) + 25O₂(gas) → 16CO₂(gas) + 18H₂O(gas). The entropy increase because there are 25 moles of gas reactants and 34 moles of gas products.

C. C₆H₁₂O₆(solid) + 6O₂(gas) → 6CO₂(gas) + 6H₂O(liquid). The entropy is the same because there are 6 moles of gas reactants and 6 moles of gas products.

D. 2C₇H₅N₃O₆(solid) → 3N₂(gas) + 7CO(gas) + 5H₂O(gas) + 7 C(solid). The entropy increase because there are no moles of gas reactants and 15 moles of gas products.

E. NaHCO₃(solid) + C₂H₄O₂(liquid) → C₂H₃NaO₂(solid)+ H₂O(liquid) + CO₂(gas). The entropy increase because there are no moles of gas reactants and there is one mole of gas product.

Thus, answer is A.

I hope it helps!