Final answer:
CH₃OH(l) has a lower standard molar entropy than CH₃OH(g) because the gaseous state allows molecules more freedom of motion, leading to a higher number of available microstates, which corresponds to higher entropy. Option A is the correct answer.
Step-by-step explanation:
When comparing CH₃OH(l) and CH₃OH(g), CH₃OH(l) has a lower standard molar entropy than CH₃OH(g). This is because the gaseous state of a substance allows for a greater freedom of motion compared to its liquid state, which translates into a higher number of microstates available.
The higher the number of microstates, the greater the entropy. In a liquid state, the molecules are closer together and their motion is more restricted than in the gaseous state where molecules have much more freedom of movement. This concept is reinforced through examples such as the comparison between gaseous ammonia (NH3) and helium (He), where NH3, being a larger molecule with more atoms, has a higher entropy due to more available microstates.
Therefore, in the case of CH₃OH, the fact that it is in a gaseous state means that the entropy is higher as the CH₃OH molecules in gas form have more translational and rotational motions possible, compared to the liquid form where the molecules are more confined. Hence, option a is the correct answer.