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

Answer:

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

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.

Answer 2

Answer:

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

Explanation:

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