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Chapter 17: Problem 84

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

Expert verified
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

01

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.
02

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.
03

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.

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Most popular questions from this chapter

For the reaction at \(298 \mathrm{~K}\), $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{~kJ}\) and \(-176.6 \mathrm{~J} / \mathrm{K}, \mathrm{re}-\) spectively. What is the value of \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) ? Assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G\) negative above or below this temperature?

The Ostwald process for the commercial production of nitric acid involves three steps: a. Calculate \(\Delta H^{\circ}, \Delta S^{\circ}, \Delta G^{\circ}\), and \(K\) (at \(298 \mathrm{~K}\) ) for each of the three steps in the Ostwald process (see Appendix 4 ). b. Calculate the equilibrium constant for the first step at \(825^{\circ} \mathrm{C}\), assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. c. Is there a thermodynamic reason for the high temperature in the first step, assuming standard conditions?

Ethanethiol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SH} ;\right.\) also called ethyl mercaptan) is commonly added to natural gas to provide the "rotten egg" smell of a gas leak. The boiling point of ethanethiol is \(35^{\circ} \mathrm{C}\) and its heat of vaporization is \(27.5 \mathrm{~kJ} / \mathrm{mol}\). What is the entropy of vaporization for this substance?

Consider the following reaction at \(298 \mathrm{~K}\) : $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)$$ An equilibrium mixture contains \(\mathrm{O}_{2}(\mathrm{~g})\) and \(\mathrm{SO}_{3}(g)\) at partial pressures of \(0.50\) atm and \(2.0\) atm, respectively. Using data from Appendix 4 , determine the equilibrium partial pressure of \(\mathrm{SO}_{2}\) in the mixture. Will this reaction be most favored at a high or a low temperature, assuming standard conditions?

Two crystalline forms of white phosphorus are known. Both forms contain \(\mathrm{P}_{4}\) molecules, but the molecules are packed together in different ways. The \(\alpha\) form is always obtained when the liquid freezes. However, below \(-76.9^{\circ} \mathrm{C}\), the \(\alpha\) form spontaneously converts to the \(\beta\) form: $$\mathrm{P}_{4}(s, \alpha) \longrightarrow \mathrm{P}_{4}(s, \beta)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\) for this process. b. Predict which form of phosphorus has the more ordered crystalline structure (has the smaller positional probability).
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