Place the following in order of decreasing standard molar entropy. NaI (s) K3PO4 (aq) NaI (aq) Place the following in order of decreasing standard molar entropy. (s) (aq) (aq) NaI (aq) > K3PO4 (aq) > NaI (s) K3PO4 (aq) > NaI (aq) > NaI (s) NaI (aq) > NaI (s) > K3PO4 (aq) NaI (s) > K3PO4 (aq) > NaI (aq) NaI (s) > NaI (aq) > K3PO4 (aq)

Question

Yoselin Pope

Chemistry

1 May, 19:18

Place the following in order of decreasing standard molar entropy. NaI (s) K3PO4 (aq) NaI (aq) Place the following in order of decreasing standard molar entropy. (s) (aq) (aq) NaI (aq) > K3PO4 (aq) > NaI (s) K3PO4 (aq) > NaI (aq) > NaI (s) NaI (aq) > NaI (s) > K3PO4 (aq) NaI (s) > K3PO4 (aq) > NaI (aq) NaI (s) > NaI (aq) > K3PO4 (aq)

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Rohan Guzman · Answer 1

We need to know the order of decreasing standard molar entropy.

The correct order (as per decreasing standard molar entropy) is: K₃PO₄ (aq) > NaI (aq) > NaI (s).

Entropy is the measure or randomness. If randomness is more, entropy is more.

Randomness of gaseous substance is more than liquid substance than solid substance.

In aqueous medium, substance dissociates and produces ions.

K₃PO₄ ionizes and produces 3 K⁺ ions and one PO₄³⁻ ion; so total 4 ions. NaI ionizes and produces one Na⁺ ion and I⁻ ion; so total 2 ions.

NaI (s) does not dissociate.

So, the correct order is K₃PO₄ (aq) > NaI (aq) > NaI (s).