Classify each of the following reactions as one of the four possible types summarized in Table 19.3: (i) spontanous at all temperatures; (ii) not spontaneous at any temperature; (iii) spontaneous at low \(T\) but not spontaneous at high \(T ;\) (iv) spontaneous at high T but not spontaneous at low \(T\). $$ \begin{array}{l} \text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{NF}_{3}(g) \\ \Delta H^{\circ}=-249 \mathrm{~kJ} ; \Delta S^{\circ}=-278 \mathrm{~J} / \mathrm{K} \\ \text { (b) } \mathrm{N}_{2}(g)+3 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NCl}_{3}(g) \\ \Delta H^{\circ}=460 \mathrm{~kJ} ; \Delta S^{\circ}=-275 \mathrm{~J} / \mathrm{K} \\ \text { (c) } \mathrm{N}_{2} \mathrm{~F}_{4}(g) \longrightarrow 2 \mathrm{NF}_{2}(g) \\ \Delta H^{\circ}=85 \mathrm{~kJ} ; \Delta S^{\circ}=198 \mathrm{~J} / \mathrm{K} \end{array} $$

Short Answer

Expert verified
In summary, reaction (a) is spontaneous at all temperatures, reaction (b) is not spontaneous at any temperature, and reaction (c) is spontaneous at high \(T\), but not spontaneous at low \(T\).

Step by step solution

01

Reaction (a)

As given, for reaction (a), \(\Delta H^{\circ} = -249 kJ\) and \(\Delta S^{\circ} = -278 J/K\). Therefore, for reaction (a), the Gibbs free energy change, \(\Delta G^{\circ}\), equation will look like: $$ \Delta G^{\circ} = (-249 kJ) - T(-278 J/K) $$ Since both \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are negative, their signs will become positive when they are multiplied, and overall \(\Delta G^{\circ} < 0\). Hence, reaction (a) is spontaneous at all temperatures.
02

Reaction (b)

As given, for reaction (b), \(\Delta H^{\circ} = 460 kJ\) and \(\Delta S^{\circ} = -275 J/K\). Therefore, for reaction (b), the Gibbs free energy change, \(\Delta G^{\circ}\), equation will look like: $$ \Delta G^{\circ} = (460 kJ) - T(-275 J/K) $$ Since \(\Delta H^{\circ}\) is positive and \(\Delta S^{\circ}\) is negative, their signs will become positive when they are multiplied, making the overall \(\Delta G^{\circ} > 0\). Hence, reaction (b) is not spontaneous at any temperature.
03

Reaction (c)

As given, for reaction (c), \(\Delta H^{\circ} = 85 kJ\) and \(\Delta S^{\circ} = 198 J/K\). Therefore, for reaction (c), the Gibbs free energy change, \(\Delta G^{\circ}\), equation will look like: $$ \Delta G^{\circ} = (85 kJ) - T(198 J/K) $$ Since \(\Delta H^{\circ}\) is positive and \(\Delta S^{\circ}\) is positive, their signs will cancel each other when they are multiplied, making the overall \(\Delta G^{\circ}\) become more negative as \(T\) increases. Hence, reaction (c) is spontaneous at high \(T\), but not spontaneous at low \(T\).

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Most popular questions from this chapter

A certain reaction has \(\Delta H^{\circ}=+20.0 \mathrm{~kJ}\) and $\Delta S^{\circ}=\( \)+100.0 \mathrm{~J} / \mathrm{K} .$ (a) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the surroundings? (c) Calculate \(\Delta G^{\circ}\) for the reaction at $298 \mathrm{~K} .(\mathbf{d})\( Is the reaction spontaneous at \)298 \mathrm{~K}$ under standard conditions?

The potassium-ion concentration in blood plasma is about $5.0 \times 10^{-3} \mathrm{M}$, whereas the concentration in muscle-cell fluid is much greater \((0.15 \mathrm{M})\). The plasma and intracellular fluid are separated by the cell membrane, which we assume is permeable only to \(\mathrm{K}^{+}\). (a) What is \(\Delta G\) for the transfer of \(1 \mathrm{~mol}\) of \(\mathrm{K}^{+}\) from blood plasma to the cellular fluid at body temperature \(37^{\circ} \mathrm{C} ?\) (b) What is the minimum amount of work that must be used to transfer this \(\mathrm{K}^{+} ?\)

Most liquids follow Trouton's rule (see Exercise 19.93 ), which states that the molar entropy of vaporization is approximately $88 \pm 5 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$. The normal boiling points and enthalpies of vaporization of several organic liquids are as follows: \begin{tabular}{lcc} \hline & Normal Boiling & \\ Substance & Point \(\left({ }^{\circ} \mathrm{C}\right)\) & $\Delta H_{\text {vap }}(\mathrm{k} / / \mathrm{mol})$ \\ \hline Acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}\) & 56.1 & 29.1 \\\ Dimethyl ether, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}\) & -24.8 & 21.5 \\\ Ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) & 78.4 & 38.6 \\ Octane, \(\mathrm{C}_{\mathrm{s}} \mathrm{H}_{18}\) & 125.6 & 34.4 \\ Pyridine, \(\mathrm{C}_{5} \mathrm{H}_{\mathrm{S}} \mathrm{N}\) & 115.3 & 35.1 \\\ \hline \end{tabular} (a) Calculate \(\Delta S_{\text {vap }}\) for each of the liquids. Do all the liquids obey Trouton's rule? (b) With reference to intermolecular forces (Section 11.2), can you explain any exceptions to the rule? (c) Would you expect water to obey Trouton's rule? By using data in Appendix \(\mathrm{B}\), check the accuracy of your conclusion. (d) Chlorobenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}\right)\) boils at \(131.8^{\circ} \mathrm{C}\). Use Trouton's rule to estimate $\Delta H_{\text {vap }}$ for this substance.

(a) Write the chemical equations that correspond to \(\Delta G_{i}^{9}\) for \(\mathrm{CH}_{4}(g)\) and for \(\mathrm{NaCl}(s) .\) (b) For these formation reactions, compare \(\Delta G_{f}^{\circ}\) and \(\Delta H_{f}\). (c) In general, under which condition is \(\Delta G\), more negative (less positive) than \(\Delta H_{f}\) ? (i) When the temperature is high, (ii) when \(\Delta S_{f}^{\circ}\) is positive, (iii) when the reaction is reversible.

Sulfur dioxide reacts with strontium oxide as follows: $$ \mathrm{SO}_{2}(g)+\mathrm{SrO}(g) \longrightarrow \mathrm{SrSO}_{3}(s) $$ (a) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ} .\) (b) If you had only standard enthalpy data for this reaction, how would you estimate the value of \(\Delta G^{\circ}\) at \(298 \mathrm{~K},\) using data from Appendix \(\mathrm{C}\) on other substances.

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