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Chapter 17: Problem 39

Given the values of \(\Delta H\) and \(\Delta S,\) which of the following changes will be spontaneous at constant \(T\) and \(P ?\) a. $\Delta H=+25 \mathrm{kJ}, \Delta S=+5.0 \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}$ b. $\Delta H=+25 \mathrm{kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}$ c. $\Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{J} / \mathrm{K}, T=298 \mathrm{K}$ d. $\Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K}$

Short Answer

Expert verified
Options b, c, and d represent spontaneous reactions at constant T and P, with ∆G values of -5000 J, -11490 J, and -2000 J, respectively.

Step by step solution

01

Recall the Gibbs free energy equation

Recall the formula for Gibbs free energy change: ∆G = ∆H - T∆S, where ∆G is Gibbs free energy change, ∆H is enthalpy change, T is temperature in Kelvin, and ∆S is entropy change. Step 2: Calculate ∆G for each option
02

Calculate ∆G for each option

Plug the given values for ∆H, ∆S, and T into the formula and calculate ∆G for each of the four cases. a) ∆H = +25 kJ, ∆S = +5.0 J/K, T = 300 K ∆G = (25*1000) - 300*(5) = 25000 - 1500 = 23500 J b) ∆H = +25 kJ, ∆S = +100 J/K, T = 300 K ∆G = (25*1000) - 300*(100) = 25000 - 30000 = -5000 J c) ∆H = -10 kJ, ∆S = +5.0 J/K, T = 298 K ∆G = (-10*1000) - 298*(5) = -10000 - 1490 = -11490 J d) ∆H = -10 kJ, ∆S = -40 J/K, T = 200 K ∆G = (-10*1000) - 200*(-40) = -10000 + 8000 = -2000 J Step 3: Determine if the reaction is spontaneous
03

Determine if the reaction is spontaneous

Analyze the calculated ∆G values to determine if the reaction is spontaneous or not. If ∆G is negative, the reaction is spontaneous. a) ∆G = 23500 J - The reaction is NOT spontaneous. b) ∆G = -5000 J - The reaction IS spontaneous. c) ∆G = -11490 J - The reaction IS spontaneous. d) ∆G = -2000 J - The reaction IS spontaneous. As a result, options b, c, and d represent spontaneous reactions at constant T and P.

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Most popular questions from this chapter

Which of the following reactions (or processes) are expected to have a negative value for \(\Delta S^{\circ} ?\) a. $\mathrm{SiF}_{6}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)+\mathrm{SiF}_{4}(g)$ b. $4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)$ c. \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{COCl}_{2}(g)\) d. $\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)$ e. \(\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\)

Two crystalline forms of white phosphorus are known. Both forms contain \(\mathrm{P}_{4}\) molecules, but the molecules are packed together in different ways. The \(\alpha\) form is always obtained when the liquid freezes. However, below \(-76.9^{\circ} \mathrm{C},\) the \(\alpha\) form spontaneously converts to the \(\beta\) form: $$\mathrm{P}_{4}(s, \alpha) \longrightarrow \mathrm{P}_{4}(s, \beta)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\) for this process. b. Predict which form of phosphorus has the more ordered crystalline structure (has the smaller positional probability).

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 \(\Delta S=+60 . \mathrm{JK}\) c. \(\Delta H=+18 \mathrm{kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) d. \(\Delta H=-18 \mathrm{kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\)

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_{\mathrm{sys}}, \Delta S_{\mathrm{surt}},\( and \)\Delta S_{\mathrm{univ}} ?$

Consider two perfectly insulated vessels. Vessel 1 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and water at \(0^{\circ} \mathrm{C}\) . Vessel 2 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and a saltwater solution at \(0^{\circ} \mathrm{C}\) . Consider the process $\mathrm{H}_{2} \mathrm{O}(s) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)$ a. Determine the sign of \(\Delta S, \Delta S_{\text { sum }}\) and $\Delta S_{\text { univ }}$ for the process in vessel 1 . b. Determine the sign of \(\Delta S, \Delta S_{\text { sum }},\) and $\Delta S_{\text { univ }}\( for the process in vessel \)2 .$ (Hint: Think about the effect that a salt has on the freezing point of a solvent.)
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