Question

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

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

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: \[\Delta S_{surr} > 0\]

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: \[\Delta S_{surr} < 0\]