Question

For the process,

$\mathit{A}\mathbf{\left(}\mathit{I}\mathbf{\right)}\mathbf{\to }\mathit{A}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

which direction is favored by changes in energy probability? Positional probability? Explain your answer. If you wanted to favor the process as written, would you raise or lower the temperature of the system? Explain.

Short answer

Forward direction will be favoured according to both energy and positional probability.

Temperature should be raised according to favour the reaction as written.

Step-by-step solution

Energy Probability

For changing a liquid to the gas, the liquid has to be heated, this heat will cause to overcome the intermolecular attractions between molecules allowing the liquid molecules to fly off into the gaseous phase. Therefore, the process of boiling is endothermic that means it requires some energy to be provided from the surroundings. When a liquid boils, the entropy change is also quite large as the molecules (gaseous) are now much freer as compared to the liquid state. When the temperature is lower than the boiling point, the enthalpy term predominates and ΔG is positive. But as temperature rises above boiling point ΔG goes from being positive to negative and the process becomes favourable. So, reaction is favourable in forward direction according to energy probability.

Positional Probability

Positional probability depends on the states of matter. Generally, it increases in going from solid to liquid to gas. Because a mole of a substance has a much smaller volume in the solid state as the molecules are close together in solid state than in the gaseous state. So, in the given example as the substance is changing from liquid to gaseous in forward direction so forward direction is favoured according to position probability.

Temperature Requirement

For a reaction to be spontaneous or favourable ΔG must be negative and entropy should increase:

$\mathbf{∆}\mathit{G}\mathbf{=}\mathbf{∆}\mathit{H}\mathbf{-}\mathit{T}\mathbf{∆}\mathit{S}$

As the reaction is endothermic so ΔH is positive so for ΔG to be negative T ΔS should be higher than ΔH so temperature should be raised to make the reaction favourable.