Question

At what temperatures will the following processes be spontaneous? a. \(\Delta H=-18 \mathrm{~kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) b. \(\Delta H=+18 \mathrm{~kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\) c. \(\Delta H=+18 \mathrm{~kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) d. \(\Delta H=-18 \mathrm{~kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\)

Short answer

The processes will be spontaneous at the following temperatures: a. \(T > 300\,\text{K}\) b. \(T < 300\,\text{K}\) c. Never spontaneous d. For any temperature \(T > 0\,\text{K}\)

Step-by-step solution

Set up the equation for the process#a.

We have the equation: \[\Delta G = \Delta H - T \Delta S = (-18 \times 10^3\,\text{J}) - T(-60\,\text{J}/\text{K})\]Now we want to find the range of temperature at which \(\Delta G < 0\).

Solve for the temperature range#a.

In order to satisfy \(\Delta G < 0\), we need the inequality: \[-18 \times 10^3\,\text{J} + 60T\text{J}/\text{K} < 0\]To find the temperature range, we will solve for T: \[T > \dfrac{18 \times 10^3}{60}\text{K}\] \[T > 300\,\text{K}\]The process will be spontaneous for temperature ranges \(T > 300\,\text{K}\). Repeat this process for the other cases. b. \(\Delta H = +18\, \text{kJ}\) and \(\Delta S = +60 . \text{J} / \text{K}\)

Set up the equation for the process#b.

We have the equation: \[\Delta G = \Delta H - T \Delta S = (18 \times 10^3\,\text{J}) - T(60\,\text{J}/\text{K})\]Now we want to find the range of temperature at which \(\Delta G < 0\).

Solve for the temperature range#b.

In order to satisfy \(\Delta G < 0\), we need the inequality: \(18 \times 10^3\,\text{J} - 60T\text{J}/\text{K} < 0\)To find the temperature range, we will solve for T: \[T < \dfrac{18 \times 10^3}{60}\text{K}\] \[T < 300\,\text{K}\]The process will be spontaneous for temperature ranges \(T < 300\,\text{K}\). c. \(\Delta H = +18\, \text{kJ}\) and \(\Delta S = -60 . \text{J} / \text{K}\)

Set up the equation for the process#c.

We have the equation: \[\Delta G = \Delta H - T \Delta S = (18 \times 10^3\,\text{J}) - T(-60\,\text{J}/\text{K})\]Now we want to find the range of temperature at which \(\Delta G < 0\).

Solve for the temperature range#c.

In order to satisfy \(\Delta G < 0\), we need the inequality: \(18 \times 10^3\,\text{J} + 60T\text{J}/\text{K} < 0\] However, since both terms are positive, it is not possible for the inequality to be satisfied. This process will never be spontaneous. d. \(\Delta H = -18\, \text{kJ}\) and \(\Delta S = +60 . \text{J} / \text{K}\)

Set up the equation for the process#d.

We have the equation: \[\Delta G = \Delta H - T \Delta S = (-18 \times 10^3\,\text{J}) - T(60\,\text{J}/\text{K})\]Now we want to find the range of temperature at which \(\Delta G < 0\).

Solve for the temperature range#d.

In order to satisfy \(\Delta G < 0\), we need the inequality: \[-18 \times 10^3\,\text{J} - 60T\text{J}/\text{K} < 0\] Since \(\Delta H\) is negative, and \(\Delta S\) is positive, the equation will always be negative for all positive values of T. Therefore, this process will always be spontaneous for any temperature \(T > 0\,\text{K}\).