ENTROPY PROBLEMS
Determine if the entropy change will be positive or negative for the following reactions:
A) (NH4)2Cr2O7(s) → Cr2O3(s) + 4 H2O(l) + CO2(g)
B) 2 H2(g) + O2(g) → 2 H2O(g)
C) PCl5 → PCl3 + Cl2(g)
A) (NH4)2Cr2O7(s) → Cr2O3(s) + 4 H2O(l) + CO2(g)
B) 2 H2(g) + O2(g) → 2 H2O(g)
C) PCl5 → PCl3 + Cl2(g)
SOLUTION
Entropy of a reaction refers to the positional probabilities for each reactant. An atom in gas phase has more options for position than the same atom in a solid phase. This is why gases have more entropy than solids.
In reactions, the positional probabilities must be compared for all the reactants to the products produced.
If the reaction involves only gases, the entropy is related to the total number of moles on either side of the reaction. A decrease in the number of moles on the product side means lower entropy. An increase in the number of moles on the product side means higher entropy.
If the reaction involves multiple phases, the production of a gas typically increases the entropy much more than any increase in moles of a liquid or solid.
Reaction A
(NH4)2Cr2O7(s) → Cr2O3(s) + 4 H2O(l) + CO2(g)
The reactant side contains only one mole where the product side has six moles produced.
In reactions, the positional probabilities must be compared for all the reactants to the products produced.
If the reaction involves only gases, the entropy is related to the total number of moles on either side of the reaction. A decrease in the number of moles on the product side means lower entropy. An increase in the number of moles on the product side means higher entropy.
If the reaction involves multiple phases, the production of a gas typically increases the entropy much more than any increase in moles of a liquid or solid.
Reaction A
(NH4)2Cr2O7(s) → Cr2O3(s) + 4 H2O(l) + CO2(g)
The reactant side contains only one mole where the product side has six moles produced.
The was also a gas produced. The change in entropy will be positive.
Reaction B
2 H2(g) + O2(g) → 2 H2O(g)
There are 3 moles on the reactant side and only 2 on the product side. The change in entropy will be negative.
Reaction C
PCl5 → PCl3 + Cl2(g)
There are more moles on the product side than on the reactant side, therefore the change in entropy will be positive.
Reaction B
2 H2(g) + O2(g) → 2 H2O(g)
There are 3 moles on the reactant side and only 2 on the product side. The change in entropy will be negative.
Reaction C
PCl5 → PCl3 + Cl2(g)
There are more moles on the product side than on the reactant side, therefore the change in entropy will be positive.
ANSWER:
Reactions A and C will have positive changes in entropy.
Reaction B will have negative changes in entropy.
Reaction B will have negative changes in entropy.
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