Consider the reaction $$2 \mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\). b. Would the reaction be more spontaneous at high or low temperatures?

Short Answer

Expert verified
∆H is negative and ∆S is negative. The reaction would be more spontaneous at low temperatures.

Step by step solution

01

Analyze the reaction and predict the signs of ∆H and ∆S

For the formation of O2 from two separate O atoms, the reaction goes from dispersed atoms to a single molecule, so the entropy ∆S should be negative. Furthermore, upon bond formation, energy is released, so the enthalpy change ∆H should be negative as well. Answer: ∆H is negative and ∆S is negative.
02

Determine the spontaneity at high or low temperatures

The spontaneity of a reaction is determined by the sign of the Gibbs free energy change (∆G), which is given by the equation: \(∆G = ∆H - T∆S\) Since both ∆H and ∆S are negative, we can infer the following information: 1. At low temperatures (T is small), the term -T∆S is small and positive, making the overall ∆G negative. This means that the reaction is spontaneous at low temperatures. 2. At high temperatures (T is large), the term -T∆S becomes larger and more positive, causing a positive or less negative ∆G. This would make the reaction less spontaneous or not spontaneous at all at high temperatures. Answer: The reaction would be more spontaneous at low temperatures.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy (∆H)
Enthalpy, represented by the symbol ∆H, is a measure of the total heat content of a system. It indicates the energy change during a reaction, particularly the heat absorbed or released at constant pressure. Understanding enthalpy is crucial because chemical reactions often involve energy changes, which can be crucial in determining spontaneity.

For instance, when a reaction releases energy, such as in the formation of a chemical bond, ∆H is negative, which is referred to as an exothermic process. This means the surroundings absorb heat from the system. On the contrary, if a reaction absorbs energy from the surroundings, ∆H is positive, and this is an endothermic process. Enthalpy changes are vital for predicting how temperature affects the spontaneity of a reaction, as seen in the given exercise.
Entropy (∆S)
Entropy, conveyed as ∆S, is a fundamental concept in chemistry that describes the degree of disorder or randomness in a system. Every substance has some intrinsic entropy, and changes in physical states, mixing of substances, or chemical reactions can alter this entropy.

The second law of thermodynamics states that the entropy of the universe tends to increase over time, which often means that processes that increase entropy are naturally favored. A positive ∆S indicates an increase in disorder—such as solid melting into liquid—while a negative ∆S, as in the formation of O_2 from separate atoms, implies a decrease in disorder. By examining the entropy change in chemical reactions, scientists can discern important aspects of the reaction's spontaneity and the conditions under which it will proceed.
Gibbs Free Energy (∆G)
Gibbs free energy, designated as ∆G, is the single most useful criterion for predicting the spontaneity of a reaction at constant temperature and pressure. It combines the concepts of enthalpy and entropy to offer a holistic view:ΔG = ΔH - TΔSWhere T is the absolute temperature in Kelvin. If ∆G is negative, the reaction is spontaneous, and if ∆G is positive, it is non-spontaneous.

For reactions where both ∆H and ∆S are negative, temperature plays a pivotal role in determining spontaneity. At lower temperatures, the reaction is more likely to be spontaneous because the entropic penalty (-TΔS) is less significant. This is why, as in the discussed exercise, certain reactions are more spontaneous at lower temperatures. Understanding how Gibbs free energy can predict the conditions under which reactions proceed is a cornerstone of chemical thermodynamics.

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Most popular questions from this chapter

For the reaction at \(298 \mathrm{~K}\), $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{~kJ}\) and \(-176.6 \mathrm{~J} / \mathrm{K}, \mathrm{re}-\) spectively. What is the value of \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) ? Assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G\) negative above or below this temperature?

a. Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{\mathrm{f}} / k_{\mathrm{r}}\), where \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) are the rate constants for the forward and reverse reactions. Hint: Use the relationship \(\Delta G^{\circ}=-R T \ln (K)\) and represent \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) using the Arrhenius equation \(\left(k=A e^{-E_{2} / R T}\right)\). b. Why is the following statement false? "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction."

Entropy has been described as "time's arrow." Interpret this view of entropy.

List three different ways to calculate the standard free energy change, \(\Delta G^{\circ}\), for a reaction at \(25^{\circ} \mathrm{C}\). How is \(\Delta G^{\circ}\) estimated at temperatures other than \(25^{\circ} \mathrm{C}\) ? What assumptions are made?

Given the following data: $$\begin{aligned}2 \mathrm{H}_{2}(g)+\mathrm{C}(s) \longrightarrow \mathrm{CH}_{4}(g) & & \Delta G^{\circ}=-51 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) & \Delta \mathrm{H}_{2} \mathrm{O}(l) & & \Delta G^{\circ}=-474 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta G^{\circ} &=-394 \mathrm{~kJ}\end{aligned}$$ Calculate \(\Delta G^{\circ}\) for \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(l) .\)

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