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Chapter 17: Problem 55

Consider the reaction $$\begin{array}{l}{2 \mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)} \\\ {\text { sof } \Delta H \text { and } \Delta S}\end{array}$$ 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
a. The signs of \(\Delta H\) and \(\Delta S\) are both negative, with \(\Delta H < 0\) and \(\Delta S < 0\). b. The reaction would be more spontaneous at low temperatures.

Step by step solution

01

Predict the sign of ∆H

To predict the sign of the change in enthalpy (∆H), we need to consider the bond energies of the reactants and products. In the given reaction, two individual oxygen molecules (O) are combining to form an O2 molecule (oxygen gas). Oxygen has a double bond, which is relatively strong. Since bond formation is an exothermic process (energy is released), we can expect that the change in enthalpy for this reaction will be negative (∆H < 0).
02

Predict the sign of ∆S

Entropy (∆S) is a measure of the disorder of a system. In this reaction, two oxygen molecules are combining to form one molecule of oxygen gas. The reaction results in a decrease in the number of particles in the system. So, the order of the system increases, and we can expect a decrease in entropy (∆S < 0).
03

Determine the spontaneity of the reaction considering temperature

According to the Gibbs free energy equation, a reaction is spontaneous if ΔG < 0, where: ΔG = ΔH - TΔS We have predicted the signs of ∆H and ∆S to be negative. As the temperature (T) increases, the term TΔS becomes more positive. Since the signs of ∆H and ΔS are both negative, and TΔS becomes more positive with increasing temperature, the value of ΔG will be less negative (or even positive) at higher temperatures. This means that the reaction is more favorable (spontaneous) at lower temperatures. In summary: a. ∆H < 0 (negative) and ∆S < 0 (negative) b. The reaction would be more spontaneous at low temperatures.

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Most popular questions from this chapter

Hydrogen cyanide is produced industrially by the following exothermic reaction: $$2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ Is the high temperature needed for thermodynamic or kinetic reasons?

For each of the following pairs, which substance has the greater value of S? a. \(\mathrm{N}_{2} \mathrm{O}(\text { at } 0 \mathrm{K})\) or He (at 10 \(\mathrm{K} )\) b. \(\mathrm{N}_{2} \mathrm{O}(g)\) (at $1 \mathrm{atm}, 25^{\circ} \mathrm{C} )\( or He(g) (at 1 atm, \)25^{\circ} \mathrm{C} )$ c. \(\mathrm{NH}_{3}(s)\) (at 196 $\mathrm{K} ) \longrightarrow \mathrm{NH}_{3}(l)(\text { at } 196 \mathrm{K})$

Consider the reaction: $$\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{S}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ for which \(\Delta H\) is \(-233 \mathrm{kJ}\) and \(\Delta S\) is $-424 \mathrm{J} / \mathrm{K}$ . a. Calculate the free energy change for the reaction \((\Delta G)\) at 393 \(\mathrm{K} .\) b. Assuming \(\Delta H\) and \(\Delta S\) do not depend on temperature, at what temperatures is this reaction spontaneous?

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?

Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does \(\mathrm{O}_{2} .\) Consider the following reactions and approximate standard free energy changes: $$\mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} \quad \Delta G^{\circ}=-70 \mathrm{kJ}$$ $$\mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} \quad \Delta G^{\circ}=-80 \mathrm{kJ} $$ Using these data, estimate the equilibrium constant value at $25^{\circ} \mathrm{C}$ for the following reaction: $$\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}$$
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