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Chapter 17: Problem 32

At what temperatures will the following processes be spontaneous? a. \(\Delta H=-18 \mathrm{~kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) b. \(\Delta H=+18 \mathrm{~kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\) c. \(\Delta H=+18 \mathrm{~kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) d. \(\Delta H=-18 \mathrm{~kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\)

Short Answer

Expert verified
The processes will be spontaneous at the following temperatures: a. \(T > 300\,\text{K}\) b. \(T < 300\,\text{K}\) c. Never spontaneous d. For any temperature \(T > 0\,\text{K}\)

Step by step solution

01

Set up the equation for the process#a.

We have the equation: \[\Delta G = \Delta H - T \Delta S = (-18 \times 10^3\,\text{J}) - T(-60\,\text{J}/\text{K})\]Now we want to find the range of temperature at which \(\Delta G < 0\).
02

Solve for the temperature range#a.

In order to satisfy \(\Delta G < 0\), we need the inequality: \[-18 \times 10^3\,\text{J} + 60T\text{J}/\text{K} < 0\]To find the temperature range, we will solve for T: \[T > \dfrac{18 \times 10^3}{60}\text{K}\] \[T > 300\,\text{K}\]The process will be spontaneous for temperature ranges \(T > 300\,\text{K}\). Repeat this process for the other cases. b. \(\Delta H = +18\, \text{kJ}\) and \(\Delta S = +60 . \text{J} / \text{K}\)
03

Set up the equation for the process#b.

We have the equation: \[\Delta G = \Delta H - T \Delta S = (18 \times 10^3\,\text{J}) - T(60\,\text{J}/\text{K})\]Now we want to find the range of temperature at which \(\Delta G < 0\).
04

Solve for the temperature range#b.

In order to satisfy \(\Delta G < 0\), we need the inequality: \(18 \times 10^3\,\text{J} - 60T\text{J}/\text{K} < 0\)To find the temperature range, we will solve for T: \[T < \dfrac{18 \times 10^3}{60}\text{K}\] \[T < 300\,\text{K}\]The process will be spontaneous for temperature ranges \(T < 300\,\text{K}\). c. \(\Delta H = +18\, \text{kJ}\) and \(\Delta S = -60 . \text{J} / \text{K}\)
05

Set up the equation for the process#c.

We have the equation: \[\Delta G = \Delta H - T \Delta S = (18 \times 10^3\,\text{J}) - T(-60\,\text{J}/\text{K})\]Now we want to find the range of temperature at which \(\Delta G < 0\).
06

Solve for the temperature range#c.

In order to satisfy \(\Delta G < 0\), we need the inequality: \(18 \times 10^3\,\text{J} + 60T\text{J}/\text{K} < 0\] However, since both terms are positive, it is not possible for the inequality to be satisfied. This process will never be spontaneous. d. \(\Delta H = -18\, \text{kJ}\) and \(\Delta S = +60 . \text{J} / \text{K}\)
07

Set up the equation for the process#d.

We have the equation: \[\Delta G = \Delta H - T \Delta S = (-18 \times 10^3\,\text{J}) - T(60\,\text{J}/\text{K})\]Now we want to find the range of temperature at which \(\Delta G < 0\).
08

Solve for the temperature range#d.

In order to satisfy \(\Delta G < 0\), we need the inequality: \[-18 \times 10^3\,\text{J} - 60T\text{J}/\text{K} < 0\] Since \(\Delta H\) is negative, and \(\Delta S\) is positive, the equation will always be negative for all positive values of T. Therefore, this process will always be spontaneous for any temperature \(T > 0\,\text{K}\).

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Most popular questions from this chapter

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

The enthalpy of vaporization of chloroform \(\left(\mathrm{CHCl}_{3}\right)\) is \(31.4\) \(\mathrm{kJ} / \mathrm{mol}\) at its boiling point \(\left(61.7^{\circ} \mathrm{C}\right) .\) Determine \(\Delta S_{\mathrm{sys}}, \Delta S_{\mathrm{sur}}\), and \(\Delta S_{\text {univ }}\) when \(1.00 \mathrm{~mol}\) chloroform is vaporized at \(61.7^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm} .\)

Consider the reaction $$2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g)$$ a. Calculate \(\Delta G^{\circ}\) for this reaction. The \(\Delta G_{\mathrm{f}}^{\circ}\) values for \(\mathrm{POCl}_{3}(g)\) and \(\mathrm{PCl}_{3}(g)\) are \(-502 \mathrm{~kJ} / \mathrm{mol}\) and \(-270 . \mathrm{kJ} / \mathrm{mol}\), respectively. b. Is this reaction spontaneous under standard conditions at \(298 \mathrm{~K} ?\) c. The value of \(\Delta S^{\circ}\) for this reaction is \(179 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). At what temperatures is this reaction spontaneous at standard conditions? Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Some water is placed in a coffee-cup calorimeter. When \(1.0 \mathrm{~g}\) of an ionic solid is added, the temperature of the solution increases from \(21.5^{\circ} \mathrm{C}\) to \(24.2^{\circ} \mathrm{C}\) as the solid dissolves. For the dissolving process, what are the signs for \(\Delta S_{\text {sys }}, \Delta S_{\text {surr }}\), and \(\Delta S_{\text {univ }}\) ?

The equilibrium constant \(K\) for the reaction $$2 \mathrm{Cl}(g) \rightleftharpoons \mathrm{Cl}_{2}(g)$$ was measured as a function of temperature (Kelvin). A graph of \(\ln (K)\) versus \(1 / T\) for this reaction gives a straight line with a slope of \(1.352 \times 10^{4} \mathrm{~K}\) and a \(y\) -intercept of \(-14.51\). Determine the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. See Exercise 71 .
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