At what temperatures will the following processes be...

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 $\Delta S=+60 . \mathrm{J} / \mathrm{K}$

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 $\Delta S=+60 . \mathrm{J} / \mathrm{K}$

Key Concepts

Critical Temperature

The concept of a critical or threshold temperature arises from setting the Gibbs free energy change to zero (?G = 0) and solving for temperature, giving T = ?H/?S. This temperature marks the point at which a process transitions from non-spontaneous to spontaneous, or vice versa.

Temperature Dependence

Temperature plays a crucial role in the spontaneity of a process because it influences the T?S term in the Gibbs free energy equation (?G = ?H - T?S). The balance between enthalpy and entropy contributions changes with temperature, affecting whether ?G becomes negative.

Entropy Change

Entropy change (?S) measures the degree of disorder or randomness in a system during a process. An increase in entropy (positive ?S) contributes to the spontaneity of a process, whereas a decrease in entropy (negative ?S) tends to oppose it.

Enthalpy Change

Enthalpy change (?H) reflects the heat exchanged during a process. An exothermic process, which has a negative ?H, tends to be spontaneous under favorable conditions, while an endothermic process (positive ?H) requires additional factors, such as temperature, to become spontaneous.

Gibbs Free Energy

Gibbs free energy is a thermodynamic function that determines if a process will occur spontaneously under constant pressure and temperature. If the change in Gibbs free energy (?G) is negative, the process proceeds spontaneously.

Transcript

00:01 So for determining the temperatures in which certain reactions will be spontaneous and given certain delta h and delta s or change in enthalpy and change in entropy respectively values.

00:15 So first we're going to look at this equation over here where we have delta g or the change in free energy equals delta h minus t, the temperature in kelvin's, times delta s.

00:30 And so given this information, we can make these conclusions that if the delta h value is negative and the delta s value is positive, then delta g will be negative for all temperatures.

00:46 And when delta g is negative, that means that the reaction is spontaneous and it's thermodynamically favorable.

00:53 So for this case, if delta g is negative, then it's spontaneous at all temperatures.

01:00 And when delta h is negative and delta s is negative, then delta g will be negative, but only for low temperatures.

01:07 So we'd have to mathematically figure that out.

01:10 And when delta h is positive and delta s is also positive, then delta g will be negative, so the reaction will be spontaneous, but only for high temperatures, which once again, we'd have to algebraically figure that out.

01:25 And when delta h is positive, delta s is negative, then our delta g will be positive for all values, which means it will not be spontaneous at any temperatures.

01:43 So let's see if we have a delta h value of positive 18 kilojoules, and we have a delta s value of negative.

02:00 60 joules per kelvin.

02:05 Since delta h, that sign is positive with a positive 18 kilojoules and delta has a negative sign at negative 60 joules per kelvin, we can determine that it will not be spontaneous at any temperature.

02:21 And so when we have a delta h that is negative, so at 18 kilojoules, and a delta s that is positive at 60 joules per degrees kelvin.

02:37 It follows this first rule up here where the delta h is negative, delta s is positive, so delta g will be negative at all temperatures, so that means it will be spontaneous at all temperatures.

02:52 So what about the cases here, when delta h and delta s have the same signs? so for that, we'd have to algebraically solve it...

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