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Chapter 17: Problem 51

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
a. For the given reaction, ΔH < 0 (exothermic) and ΔS < 0 (decrease in disorder). b. The reaction would be more spontaneous at low temperatures, as ΔG would be more negative in this case.

Step by step solution

01

a. Predict the signs of ΔH and ΔS

In this reaction, we have two oxygen atoms combining to form an oxygen molecule. We can make predictions about ΔH and ΔS based on the characteristics of the reactants and products. ΔH: The formation of chemical bonds typically releases heat, making the reaction exothermic. As the oxygen atoms form an oxygen molecule, they are forming a bond, so we would expect the enthalpy change to be negative (exothermic). Therefore, ΔH < 0. ΔS: Entropy is a measure of the randomness or disorder of a system. As two oxygen atoms combine to form a single molecule, the system becomes more ordered and less random. As a result, the entropy change would be negative. Therefore, ΔS < 0.
02

b. Spontaneity at High or Low Temperatures

To determine whether the reaction is more spontaneous at high or low temperatures, we will consider the Gibbs free energy change equation: ΔG = ΔH - TΔS Where ΔG is the Gibbs free energy change, T is the temperature in Kelvin, and we have already determined that ΔH < 0 and ΔS < 0. When ΔG is negative, the reaction is spontaneous. Since both ΔH and ΔS are negative, the equation becomes: ΔG = Negative - T(Negative) At low temperatures, the TΔS term will be smaller in magnitude, making the overall ΔG more negative. When the temperature is increased, the TΔS term will become larger in magnitude, making ΔG less negative or even positive. Thus, the reaction would be more spontaneous at low temperatures.

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Most popular questions from this chapter

The synthesis of glucose directly from \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) and the synthesis of proteins directly from amino acids are both nonspontaneous processes under standard conditions. Yet it is necessary for these to occur for life to exist. In light of the second law of thermodynamics, how can life exist?

The Ostwald process for the commercial production of nitric acid involves three steps: $$ \begin{array}{l} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \underset{825^{\circ} \mathrm{C}}{\longrightarrow} 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(l)+\mathrm{NO}(g) \end{array} $$ a. Calculate \(\Delta H^{\circ}, \Delta S^{\circ}, \Delta G^{\circ}\), and \(K\) (at \(\left.298 \mathrm{~K}\right)\) for each of the three steps in the Ostwald process (see Appendix 4). b. Calculate the equilibrium constant for the first step at \(825^{\circ} \mathrm{C}\), assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. c. Is there a thermodynamic reason for the high temperature in the first step, assuming standard conditions?

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 \(\left(k=A e^{-E_{\mathrm{a}} / R T}\right) .\) 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."

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: $$ \begin{array}{cl} \mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\ \mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{array} $$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$ \mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2} $$

In the text, the equation $$ \Delta G=\Delta G^{\circ}+R T \ln (Q) $$ was derived for gaseous reactions where the quantities in \(Q\) were expressed in units of pressure. We also can use units of \(\mathrm{mol} / \mathrm{L}\) for the quantities in \(Q\), specifically for aqueous reactions. With this in mind, consider the reaction $$ \mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q) $$ for which \(K_{\mathrm{a}}=7.2 \times 10^{-4}\) at \(25^{\circ} \mathrm{C} .\) Calculate \(\Delta G\) for the reaction under the following conditions at \(25^{\circ} \mathrm{C}\). a. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \mathrm{M}\) b. \([\mathrm{HF}]=0.98 M,\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=2.7 \times 10^{-2} M\) c. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-5} \mathrm{M}\) d. \([\mathrm{HF}]=\left[\mathrm{F}^{-}\right]=0.27 M,\left[\mathrm{H}^{+}\right]=7.2 \times 10^{-4} M\) e. \([\mathrm{HF}]=0.52 M,\left[\mathrm{~F}^{-}\right]=0.67 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-3} M\) Based on the calculated \(\Delta G\) values, in what direction will the reaction shift to reach equilibrium for each of the five sets of conditions?
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