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Predict in which of the following entropy decreases:
A. A liquid crystallizes into a solid.
B. Temperature of a crystalline solid raised from 0K to 115K
C. 2NaHCO2(s)Na2CO3(s)+CO2(g)+H2O(g)
D. H2(g)2H(g)

Answer
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Hint: When a small amount of heat ΔQ is added to a substance at temperature T, without changing its temperature appreciably, the entropy of the substance changes by When heat is removed, the entropy decreases, when heat is added the entropy increases. Entropy has units of Joules per Kelvin.

Complete step by step answer:
(a) Option 1st: When a liquid crystallizes into a solid, the molecule obtains an ordered state and hence, entropy decreases.
 (b) Option 2nd: At 0K, the constituent particles are static and entropy is minimum. If temperature is raised to 115K, these begin to move and oscillate about their equilibrium positions in the lattice and the system becomes more disordered, hence entropy increases.
(c) Option 3rd:
2NaHCO2(s)Na2CO3(s)+CO2(g)+H2O(g)
In this reaction, the reactant, NaHCO3 is a solid and it has low entropy. Among products, there are one solid and two gases. So, the products represent a condition of higher entropy.
(d) Option 4th:
H2(g)2H(g)
Here, one molecule of hydrogen gives two hydrogen atoms i.e. the number of particles increases leading to more disordered state. Two moles of H-atoms have higher entropy than one mole of hydrogen molecule.

So the correct option is A

Note:
A phase change from a liquid to a solid (i.e. freezing), or from a gas to a liquid (i.e. condensation) results in decrease in the disorder of the substance, and a decrease in the entropy. The entropy of a framework relies upon its inward energy and its outer boundaries, for example, its volume. In as far as possible, this reality prompts a condition relating the adjustment in the inward energy U to changes in the entropy and the outside boundaries.


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