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Question on entropy changes

There’s a multiple choice question on entropy which asks ‘in which one of the following reactions is there a decrease in entropy?’
A. [Fe(H2O)6]³⁺ (aq)+3C2O4²⁻ (aq) -> [fe(C2O4)3]³ (aq) + 6H20 (l)
B. [Cu(H2O)6]² (aq) + EDTA⁴ (aq) -> [Cu(EDTA)]² (aq) + 6H2O (l)
C. [CoCl4]²⁻ (aq) + 6H2O(l) -> [Co(H2O)6]²⁺ (aq) + 4Cl⁻ (aq)
D. Na2CO3 (s) + 2H⁺ (aq) -> 2Na⁺ (aq) + CO2(g) + H2O(l)

I don’t understand why the answer is D and not C, I thought that D would have an increase in entropy because there are more moles in the products and it goes from s + aq -> aq + g + l, I thought the products would have a higher entropy than s + aq. Also C has more moles in reactants than products so surely there would be a decrease? I would really appreciate it if anyone could explain this to me
Thanks in advance
Gases have much more entropy than other states. C has 1 aqueous and 6 water going to 5 aqueous. Having water in addition to aqueous solution has very little change in entropy. Essentially how many ways can the 1 aqueous solute be arranged in solution vs same but with solution with 6 extra water.
(edited 11 months ago)
Reply 2
Original post by Old man1234
Gases have much more entropy than other states. C has 1 aqueous and 6 water going to 5 aqueous. Having water in addition to aqueous solution has very little change in entropy. Essentially how many ways can the 1 aqueous solute be arranged in solution vs same but with solution with 6 extra water

But if gases have much more entropy then wouldn’t that be an increase in entropy? And if there is 6 extra water, does that mean there are more ways the molecules can be arrange compared to the solid? So wouldn’t that mean there would be more entropy in the reactants than products for C?
I think C is definitely an increase. But now I'm confused about d.
D is definitely an increased. Sorry I'm no help.
I think it should be A.

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