Question

In each of the following pairs, choose the substance with a lower entropy. (a) \(\mathrm{H}_{2} \mathrm{O}(l)\) at \(10^{\circ} \mathrm{C}, \mathrm{H}_{2} \mathrm{O}(l)\) at \(30^{\circ} \mathrm{C}\) (b) \(\mathrm{C}\) (graphite), \(\mathrm{C}\) (diamond) (c) \(\mathrm{Cl}_{2}(l), \mathrm{Cl}_{2}(g),\) both at room temperature

Short answer

Answer: (a) H2O(l) at 10°C, (b) diamond, and (c) Cl2(l).

Step-by-step solution

Pair (a) Comparison: H2O(l) at 10°C and 30°C

In this pair, the substance is the same, but the temperatures are different. Entropy increases as the temperature of a substance increases. Since 30°C is greater than 10°C, the H2O(l) at 30°C has a higher entropy than the H2O(l) at 10°C. Therefore, H2O(l) at 10°C has the lower entropy.

Pair (b) Comparison: Carbon in graphite form and diamond form

In this pair, both substances are composed of carbon atoms but arranged in different structures. Graphite is composed of loosely connected layers and has a more disordered structure, while carbon atoms in diamond are organized in a tightly packed and orderly structure. As a result, graphite has higher entropy due to its more disordered arrangement. Therefore, diamond has the lower entropy.

Pair (c) Comparison: Cl2(l) and Cl2(g) at room temperature

In this pair, the same substance (Cl2) is in two different phases (liquid and gas) at room temperature. Entropy increases with the degree of freedom or movement of particles in a substance. Gas particles have more freedom to move as compared to liquid particles. Thus, Cl2(g) has a higher entropy due to the increased freedom of its particles. Therefore, Cl2(l) has the lower entropy.