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Chapter 16: Problem 33

Predict the sign of \(\Delta S_{\text {surr }}\) for the following processes. a. \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\) b. \(\mathrm{I}_{2}(g) \longrightarrow \mathrm{I}_{2}(s)\)

Short Answer

Expert verified
For the processes: a. \(H_2O(l) \longrightarrow H_2O(g)\), the entropy change for the surroundings is positive. Therefore, \(\Delta S_{surr} > 0\). b. \(I_2(g) \longrightarrow I_2(s)\), the entropy change for the surroundings is negative. Therefore, \(\Delta S_{surr} < 0\).

Step by step solution

01

a. H2O(l) → H2O(g)

For the process: \(H_2O(l) \longrightarrow H_2O(g)\) The phase transition is from a liquid to a gas. Gaseous molecules have greater mobility and more possible arrangements than liquid molecules. As energy is transferred from the system to the surroundings, the disorder and randomness increase in the surroundings. As a result, the entropy change for the surroundings is positive. So, the sign of ΔSsurr in this case is: \[\Delta S_{surr} > 0\]
02

b. I2(g) → I2(s)

For the process: \(I_2(g) \longrightarrow I_2(s)\) The phase transition is from a gas to a solid. Solid molecules have less mobility and fewer possible arrangements than gaseous molecules. As energy is transferred from the surroundings to the system, the disorder and randomness decrease in the surroundings. Therefore, the entropy change for the surroundings is negative. So, the sign of ΔSsurr in this case is: \[\Delta S_{surr} < 0\]

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Most popular questions from this chapter

List three different ways to calculate the standard free energy change, \(\Delta G^{\circ},\) for a reaction at \(25^{\circ} \mathrm{C} .\) How is \(\Delta G^{\circ}\) estimated at temperatures other than \(25^{\circ} \mathrm{C} ?\) What assumptions are made?

Given the following data: $$2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow 12 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\Delta G^{\circ}=-6399 \mathrm{kJ}$$ $$\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta G^{\circ}=-394 \mathrm{kJ}$$ $$\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta G^{\circ}=-237 \mathrm{kJ}$$ calculate \(\Delta G^{\circ}\) for the reaction $$6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)$$

The third law of thermodynamics states that the entropy of a perfect crystal at \(0 \mathrm{K}\) is zero. In Appendix \(4, \mathrm{F}^{-}(a q), \mathrm{OH}^{-}(a q)\) and \(\mathrm{S}^{2-}(a q)\) all have negative standard entropy values. How can \(S^{\circ}\) values be less than zero?

For each of the following pairs of substances, which substance has the greater value of \(S^{\circ} ?\) a. \(\mathrm{C}_{\text {graphite }}(s)\) or \(\mathrm{C}_{\text {diamond }}(s)\) b. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\) c. \(\mathrm{CO}_{2}(s)\) or \(\mathrm{CO}_{2}(g)\)

Hydrogen sulfide can be removed from natural gas by the reaction $$2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ Calculate \(\Delta G^{\circ}\) and \(K\) (at 298 K) for this reaction. Would this reaction be favored at a high or low temperature?
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