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
a. For the process \(H_2O(l) \rightarrow H_2O(g)\), the sign of \(ΔS_{surr}\) is positive: \(ΔS_{surr}(a) > 0\). b. For the process \(I_2(g) \rightarrow I_2(s)\), the sign of \(ΔS_{surr}\) is negative: \(ΔS_{surr}(b) < 0\).

Step by step solution

01

Case (a): Conversion of liquid water to water vapor

When liquid water is converted into water vapor, the molecules gain kinetic energy and occupy a larger volume. This increase in volume means that there is more positional freedom for the water molecules, creating more disorder. Since this process increases the disorder, it means that the change in entropy of the surroundings will be positive. So the sign of \(ΔS_{surr}\) for the process \(H_2O(l) \rightarrow H_2O(g)\) is: \(ΔS_{surr}(a) > 0\)
02

Case (b): Conversion of iodine gas to solid iodine

When iodine gas is converted into solid iodine, the molecules lose kinetic energy and become more ordered, as they are now in a solid-state with a fixed arrangement. Since the process leads to a more ordered state, it means that the change in entropy of the surroundings will be negative. So the sign of \(ΔS_{surr}\) for the process \(I_2(g) \rightarrow I_2(s)\) is: \(ΔS_{surr}(b) < 0\) In summary, the signs of \(ΔS_{surr}\) for the given processes are: a. \(ΔS_{surr}(a) > 0\) for \(H_2O(l) \rightarrow H_2O(g)\) b. \(ΔS_{surr}(b) < 0\) for \(I_2(g) \rightarrow I_2(s)\)

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