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Chapter 17: Problem 90

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_{\mathrm{sys}}, \Delta S_{\mathrm{surt}},\( and \)\Delta S_{\mathrm{univ}} ?$

Short Answer

Expert verified
In the dissolving process, the signs for the entropy changes are as follows: \( \Delta S_{sys} \) (system) is positive since entropy increases with increasing disorder. \( \Delta S_{surr} \) (surroundings) is positive since heat is released to the surroundings during the exothermic process, and temperature is always positive. Finally, \( \Delta S_{univ} \) (universe) is positive because it is the sum of the positive entropy changes for the system and surroundings.

Step by step solution

01

Determine the sign of ∆S_sys

To determine the sign of ∆S_sys, we must analyze the molecular states involved in the dissolving process. The solid ionic compound is in a highly ordered state. When it dissolves in the water, the ions disperse, and the overall system becomes less ordered. The change in entropy of the system (∆S_sys) will be positive, since entropy increases with increasing disorder.
02

Determine the sign of ∆S_surr

Since the dissolving process is exothermic, it releases heat to the surroundings. The sign of ∆S_surr depends on the sign of the heat (q) released and the temperature (T) at which the process occurs. Here, the heat is released to the surroundings, making q a negative value. The temperature is always positive. Therefore, ∆S_surr = -q/T, and the negative sign of q changes to a positive sign when multiplied by the constant negative sign. So, the change in entropy of the surroundings (∆S_surr) is positive.
03

Determine the sign of ∆S_univ

To find the sign of ∆S_univ, we sum the change in entropy for the system and the surroundings. Since both ∆S_sys and ∆S_surr are positive, their sum (∆S_univ) will also be positive. In conclusion, the signs for the entropy changes are as follows: ∆S_sys: Positive ∆S_surr: Positive ∆S_univ: Positive

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Most popular questions from this chapter

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?

Consider the following reaction at \(35^{\circ} \mathrm{C} :\) $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \Delta G^{\circ}=20 . \mathrm{kJ} $$ If 2.0 atm of NOCl are reacted in a rigid container at $35^{\circ} \mathrm{C},$ calculate the equilibrium partial pressure of NO.

The equilibrium constant for a certain reaction increases by a factor of 6.67 when the temperature is increased from 300.0 \(\mathrm{K}\) to 350.0 \(\mathrm{K}\) . Calculate the standard change in enthalpy $\left(\Delta H^{\circ}\right)\( for this reaction (assuming \)\Delta H^{\circ}$ is temperature-independent).

Which of the following processes are spontaneous? a. Salt dissolves in \(\mathrm{H}_{2} \mathrm{O}\) . b. A clear solution becomes a uniform color after a few drops of dye are added. c. Iron rusts. d. You clean your bedroom.

Gas \(A_{2}\) reacts with gas \(B_{2}\) to form gas AB at a constant temperature. The bond energy of \(A B\) is much greater than that of either reactant. What can be said about the sign of \(\Delta H ? \Delta S_{\mathrm{surr}} ?\) $\Delta S$? Explain how potential energy changes for this process. Explain how random kinetic energy changes during the process.
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