Question

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 \(\Delta S_{\text {sys }}, \Delta S_{\text {surr }}\), and \(\Delta S_{\text {univ }}\) ?

Short answer

In conclusion, for the dissolving process, the signs for the entropy changes are: ΔS_sys is positive because the ionic solid dissolves into its respective ions, increasing randomness in the system, ΔS_surr is positive because the exothermic process releases heat into the surroundings, causing an increase in randomness, and ΔS_univ is positive because it is the sum of the positive ΔS_sys and ΔS_surr values.

Step-by-step solution

1. Determine the sign for ΔS_sys

To determine the sign for ΔS_sys, consider the dissolving process of the solid. Since it dissolves into its respective ions in the solution, the overall number of particles increases, leading to an increase in the randomness of the system. Therefore, the entropy of the system (ΔS_sys) is positive.

2. Determine the sign for ΔS_surr

Now we need to determine the sign of ΔS_surr. We know that the dissolving process is exothermic, which means it releases heat into the surroundings. The released heat causes the temperature of the surroundings to rise and increase the randomness of the surroundings. Therefore, the entropy of the surroundings (ΔS_surr) is positive.

3. Determine the sign for ΔS_univ

Since both ΔS_sys and ΔS_surr are positive, the change in entropy of the universe (ΔS_univ) will be the sum of the two: \(ΔS_{univ} = ΔS_{sys} + ΔS_{surr}\) As both terms are positive, the overall change in entropy of the universe (ΔS_univ) will also be positive. In conclusion, the signs for the entropy changes are: ΔS_sys is positive, ΔS_surr is positive, and ΔS_univ is positive.