Predict the conditions (high temperature, low temperature,...

Predict the conditions (high temperature, low temperature, all temperatures, or no temperatures) under which each reaction is spontaneous. a. $\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$ b. $\mathrm{CO}_{2}(s) \longrightarrow \mathrm{CO}_{2}(g)$ c. $\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g)$ d. $2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$ (endothermic)

Predict the conditions (high temperature, low temperature, all temperatures, or no temperatures) under which each reaction is spontaneous.
a. $\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$
b. $\mathrm{CO}_{2}(s) \longrightarrow \mathrm{CO}_{2}(g)$
c. $\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g)$
d. $2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$ (endothermic)

Key Concepts

Gibbs Free Energy

Gibbs free energy (?G) is the thermodynamic potential that indicates whether a process is spontaneous. A reaction is spontaneous if ?G is negative. The value of ?G is determined by the balance between the enthalpy change (?H) and the entropy change (?S) of the system, according to the relation ?G = ?H - T?S.

Enthalpy and Entropy

Enthalpy (?H) represents the heat absorbed or released during a reaction and is used to classify reactions as exothermic or endothermic. Entropy (?S) measures the disorder or randomness in a system. The interplay between ?H and ?S helps predict spontaneity; for instance, an exothermic process with increasing entropy is generally spontaneous at all temperatures, while an endothermic reaction with increasing entropy may require high temperatures to be spontaneous.

Temperature Dependence of Spontaneity

The temperature (T) term in the Gibbs free energy equation plays a crucial role in determining the sign of ?G. In cases where the reaction is endothermic (positive ?H) and increases entropy (positive ?S), a higher temperature may be necessary to achieve spontaneity because the T?S term must overcome the enthalpic cost. Conversely, for reactions that are exothermic (negative ?H) and decrease entropy (negative ?S), lower temperatures are more favorable for spontaneity.

Transcript

00:01 So this question is wanting you to figure out at what temperature range are these reactions spontaneous.

00:08 So for question, part a for this reaction, we can calculate delta h and delta s of reaction, understand their state, and see what delta g would be, and now determine what our temperature range can be.

00:26 So you can calculate delta h by doing delta h of formation of product.

00:32 Minus reactants.

00:34 And so you can find these in your textbook.

00:37 So for the liquid, it's 285 .8 minus negative 241 .8 for the gas, and that gives you minus 44.

01:00 So it gives you a negative number.

01:05 Then you can calculate delta s if you want to, but you can also just look at it intuitively and see that you're going from a gas to a liquid.

01:11 So your change in entropy is going to decrease because you're becoming more ordered.

01:18 So if your delta h is negative and your delta s is negative, delta g, which equals delta h minus t delta s, you have, and you want delta g to be as negative as possible.

01:36 So, delta h is negative and delta s is negative, which means this number overall is going to be a positive number.

01:44 So you want t to be as small as possible, so this delta h is as negative as is larger than this value.

01:52 So you can be more negative.

01:54 So this is spontaneous at low temperature, at low t.

02:06 So for the next one, calculate delta h a very similar way.

02:12 You can find these in your textbook or online.

02:18 You're just doing products minus reactants again...

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