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Chapter 17: Problem 36

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 under the following conditions: a. \(T > 300\mathrm{~K}\) b. \(T < 300\mathrm{~K}\) c. Never spontaneous d. Always spontaneous

Step by step solution

01

Case a: ΔH = -18 kJ, ΔS = -60 J/K

Let's solve for T when \( ΔG = ΔH - TΔS < 0 \): \[ -18\mathrm{~kJ} - T(-60 . \mathrm{J} / \mathrm{K}) < 0 \] First, convert ∆H to J: \[ -18000\mathrm{~J} + 60T < 0 \] Solve for T: \[ T > \frac{18000}{60} \Rightarrow T > 300\mathrm{~K} \] So, in this case, the process is spontaneous when the temperature is greater than 300 K.
02

Case b: ΔH = +18 kJ, ΔS = +60 J/K

Let's solve for T when \( ΔG = ΔH - TΔS < 0 \): \[ 18\mathrm{~kJ} - T(60 . \mathrm{J} / \mathrm{K}) < 0 \] Convert ∆H to J: \[ 18000\mathrm{~J} - 60T < 0 \] Solve for T: \[ T < \frac{18000}{60} \Rightarrow T < 300\mathrm{~K} \] In this case, the process is spontaneous when the temperature is less than 300 K.
03

Case c: ΔH = +18 kJ, ΔS = -60 J/K

Let's solve for T when \( ΔG = ΔH - TΔS < 0 \): \[ 18\mathrm{~kJ} - T(-60 . \mathrm{J} / \mathrm{K}) < 0 \] Convert ∆H to J: \[ 18000\mathrm{~J} + 60T < 0 \] Solve for T: \[ T < -\frac{18000}{60} \Rightarrow T < -300\mathrm{~K} \] However, temperature cannot be negative. Therefore, the process is never spontaneous in this case.
04

Case d: ΔH = -18 kJ, ΔS = +60 J/K

Let's solve for T when \( ΔG = ΔH - TΔS < 0 \): \[ -18\mathrm{~kJ} - T(60 . \mathrm{J} / \mathrm{K}) < 0 \] Convert ∆H to J: \[ -18000\mathrm{~J} - 60T < 0 \] Solve for T: \[ T > -\frac{-18000}{60} \Rightarrow T > -300\mathrm{~K} \] Since absolute temperature is always positive, the process is always spontaneous in this case.

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Most popular questions from this chapter

For the process \(\mathrm{A}(l) \longrightarrow \mathrm{A}(\mathrm{g})\), which direction is favored by changes in energy probability? Positional probability? Explain your answers. If you wanted to favor the process as written, would you raise or lower the temperature of the system? Explain.

The following reaction occurs in pure water: $$ \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ which is often abbreviated as $$ \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ For this reaction, \(\Delta G^{\circ}=79.9 \mathrm{~kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C}\). Calculate the value of \(\Delta G\) for this reaction at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{OH}^{-}\right]=0.15 \mathrm{M}\) and \(\left[\mathrm{H}^{+}\right]=0.71 M\).

Is \(\Delta S_{\text {surr }}\) favorable or unfavorable for exothermic reactions? Endothermic reactions? Explain.

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

Calculate \(\Delta G^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)\) at \(600 . \mathrm{K}\), using the following data: \(\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6}\) at 600. \(\mathrm{K}\) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37}\) at \(600 . \mathrm{K}\)
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