Question

Consider the following equation.

Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O(g)

H = 98.8 kJ, and S = 141.5 J/K. Is this reaction spontaneous or nonspontaneous at high and low temperatures?
A. spontaneous at high temperatures, non-spontaneous at low temperatures
B. Non-spontaneous at high and low temperatures
C. Spontaneous at low temperatures, non-spontaneous at high temperatures
D. Spontaneous at high and low temperatures

Answer 1

A.) spontaneous at high temperatures, non-spontaneous at low temperatures

Step-by-step explanation:

Edge2021

Answer 2

This reaction is A. Spontaneous at high temperatures, and non-spontaneous at low temperatures.

Step-by-step explanation:

Both the enthalpy change
`\Delta H` and the entropy change
`\Delta S` due to this reaction are positive. A chemical reaction will be spontaneous only if the change in its Gibbs Free Energy
`\Delta G = \Delta H - T \cdot \Delta S` is negative.
`T` is the absolute temperature in degrees Kelvins.

Assume that both
`\Delta H` and
`\Delta S` doesn't change much as
`T` increases. The value of
`\Delta G` will initially be close to
`\Delta H` when
`T` is small. The sign of
`\Delta G` will depends on that of
`\Delta H`. However,
`\Delta H` is positive, so at low temperatures
`\Delta G` will be positive and the reaction will be non-spontaneous.

However, as
`T` increases, the role of entropy change becomes more significant. The sign of
`\Delta G` will eventually be the opposite of
`\Delta S`. The value of
`\Delta G` will eventually drop below zero after the value of
`T` rises above
`\Delta H / \Delta S`. The reaction will eventually become spontaneous.