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Chapter 17: Problem 20

The melting point for carbon diselenide \(\left(\mathrm{CSe}_{2}\right)\) is \(-46^{\circ} \mathrm{C}\) . At a temperature of \(-75^{\circ} \mathrm{C},\) predict the signs for \(\Delta S_{\mathrm{surr}}\) and $\Delta S_{\mathrm{univ}}\( for the following process: \)\operatorname{CSe}_{2}(l) \rightarrow \operatorname{CSe}_{2}(s)$

Short Answer

Expert verified
For the process of CSe2 changing from liquid to solid at -75°C, the signs for the change in entropy are: \(\Delta S_{sys}\) is negative, \(\Delta S_{surr}\) is positive, and \(\Delta S_{univ}\) is positive.

Step by step solution

01

Determine the entropy change for the system (\(\Delta S_{sys}\))

Since we are given that CSe2 is changing from liquid to solid phase, this process is a freezing (or crystallization) process. In general, the entropy of a system decreases when a substance freezes. Therefore, for this process, \(\Delta S_{sys}\) will be negative.
02

Determine the temperature at which the phase change occurs

We are given that the phase change occurs at a temperature of -75°C. To understand the impacts on the surrounding entropy, it's useful to compare this to the melting point of the substance. The melting point for CSe2 is -46°C, which means that the temperature is lower than the melting point when the phase change occurs in this scenario.
03

Determine the sign of the entropy change for the surroundings (\(\Delta S_{surr}\))

To change from liquid to solid, heat must be released to the surroundings. This means that the heat flow is exothermic, which corresponds to a negative \(\Delta H\). Since the phase change is occurring below the melting point, the surroundings are colder, resulting in a greater impact of heat from the system. In this situation, the equation \(\Delta S_{surr} = -\frac{\Delta H}{T}\) will have a greater negative number in the numerator and a smaller absolute value (but still negative) in the denominator, resulting in a positive \(\Delta S_{surr}\).
04

Determine the sign of the entropy change for the universe (\(\Delta S_{univ}\))

The change in entropy for the universe is calculated as the sum of the entropy changes for the system and the surroundings: \(\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr}\). From Step 1, we found that \(\Delta S_{sys}\) is negative, and from Step 3, we found that \(\Delta S_{surr}\) is positive. In this case, since the process is occurring at a temperature below the melting point, the positive \(\Delta S_{surr}\) will have a greater magnitude than the negative \(\Delta S_{sys}\). Therefore, the overall \(\Delta S_{univ}\) will be positive. In conclusion, for the given process of CSe2 changing from liquid to solid at -75°C, the signs for the change in entropy are: - \(\Delta S_{sys}\): negative - \(\Delta S_{surr}\): positive - \(\Delta S_{univ}\): positive

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Most popular questions from this chapter

Consider the reaction: $$\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)$$ At \(25^{\circ} \mathrm{C}, \Delta G^{\circ}=-92.50 \mathrm{kJ}\) Which of the following statements is (are) true? a. This is an endothermic reaction. b. \(\Delta S^{\circ}\) for this reaction is negative. c. If the temperature is increased, the ratio \(\frac{\mathrm{PCl}_{5}}{\mathrm{PCl}_{3}}\) will increase. d. \(\Delta G^{\circ}\) for this reaction has to be negative at all temperatures. e. When \(\Delta G^{\circ}\) for this reaction is negative, then \(K\) is greater than 1.00 .

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. \(\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\)

For ammonia \(\left(\mathrm{NH}_{3}\right),\) the enthalpy of fusion is 5.65 \(\mathrm{kJ} / \mathrm{mol}\) and the entropy of fusion is 28.9 $\mathrm{J} / \mathrm{K} \cdot$ mol. a. Will \(\mathrm{NH}_{3}(s)\) spontaneously melt at \(200 . \mathrm{K}\) ? b. What is the approximate melting point of ammonia?

For the reaction at 298 K, $$2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are $-58.03 \mathrm{kJ}\( and \)-176.6 \mathrm{J} / \mathrm{K}$ , respectively. What is the value of \(\Delta G^{\circ}\) at 298 \(\mathrm{K}\) ? Assuming that \(\Delta H^{p}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G^{\circ}\) negative above or below this temperature?

Consider the following equilibrium constant versus temperature data for some reaction: $$\begin{array}{ll} {\boldsymbol{T}\left(^{\circ} \mathbf{C}\right)} & \quad {\text { K }} \\ \hline {109} & {2.54 \times 10^{4}} \\ {225} & {5.04 \times 10^{2}} \\ {303} & {6.33 \times 10^{1}} \\ {412} & {2.25 \times 10^{-1}} \\\ {539} & {3.03 \times 10^{-3}}\end{array}$$ Predict the signs for \(\Delta G^{\circ}, \Delta H^{\circ},\) and $\Delta S^{\circ}$ for this reaction at \(25^{\circ} \mathrm{C}\) . Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.
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