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Chapter 16: Problem 42

Predict the sign of \(\Delta S^{\circ}\) for each of the following changes. a. \(\mathrm{K}(s)+\frac{1}{2} \mathrm{Br}_{2}(g) \longrightarrow \mathrm{KBr}(s)\) b. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) c. \(\mathrm{KBr}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{Br}^{-}(a q)\) d. \(\mathrm{KBr}(s) \longrightarrow \mathrm{KBr}(l)\)

Short Answer

Expert verified
a. \(\Delta S^{\circ} < 0\), since the reaction leads to a decrease in gaseous molecules. b. \(\Delta S^{\circ} < 0\), since the reaction results in a decrease in the number of gaseous molecules. c. \(\Delta S^{\circ} > 0\), as the solid KBr dissolves, leading to an increase in entropy. d. \(\Delta S^{\circ} > 0\), as the solid KBr melts to form a liquid, increasing entropy.

Step by step solution

01

Examine the physical states of reactants and products

Here, we have a solid potassium (K) reacting with gaseous bromine (Br2) to form a solid potassium bromide (KBr).
02

Determine the change in entropy

The reactant side has one molecule of gas (Br2), while the product side has no gaseous molecules. The reaction results in a decrease in the number of gaseous molecules, and as gases have higher entropy than solids, this will result in a decrease in entropy. Therefore, the sign of \(\Delta S^{\circ}\) for this reaction will be negative (\(\Delta S^{\circ} < 0\)). #b. N2(g) + 3 H2(g) → 2 NH3(g)#
03

Examine the physical states and number of molecules

In this reaction, gaseous nitrogen (N2) reacts with gaseous hydrogen (H2) to form gaseous ammonia (NH3).
04

Determine the change in entropy

There are 4 gaseous molecules on the reactant side (1 N2 and 3 H2) and only 2 gaseous molecules on the product side (2 NH3). This reaction results in a decrease in the number of gaseous molecules going from reactants to products, which indicates a decrease in entropy. Thus, the sign of \(\Delta S^{\circ}\) for this reaction will be negative (\(\Delta S^{\circ} < 0\)). #c. KBr(s) → K+(aq) + Br-(aq)#
05

Examine the physical states and number of particles

In this reaction, solid potassium bromide (KBr) is dissolving to form potassium ions (K+) and bromide ions (Br-) in an aqueous solution.
06

Determine the change in entropy

Going from a solid to an aqueous solution leads to an increase in entropy as the ions become more randomly dispersed in the solution compared to their ordered arrangement in the solid. Therefore, the sign of \(\Delta S^{\circ}\) for this reaction will be positive (\(\Delta S^{\circ} > 0\)). #d. KBr(s) → KBr(l)
07

Examine the physical states of reactants and products

In this reaction, solid potassium bromide (KBr) is melting to form liquid potassium bromide (KBr).
08

Determine the change in entropy

The entropy of a substance increases as it transitions from solid to liquid. In this case, the reaction involves the melting of a solid, which results in an entropy increase due to the increased freedom of motion of particles in the liquid state compared to the solid state. Thus, the sign of \(\Delta S^{\circ}\) for this reaction will be positive (\(\Delta S^{\circ} > 0\)).

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Most popular questions from this chapter

Consider two perfectly insulated vessels. Vessel 1 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and water at \(0^{\circ} \mathrm{C}\). Vessel 2 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and a saltwater solution at \(0^{\circ} \mathrm{C}\). Consider the process \(\mathrm{H}_{2} \mathrm{O}(s) \rightarrow \mathrm{H}_{2} \mathrm{O}(l).\) a. Determine the sign of \(\Delta S, \Delta S_{\text {sur, }}\) and \(\Delta S_{\text {univ }}\) for the process in vessel 1. b. Determine the sign of \(\Delta S, \Delta S_{\text {sur, }}\) and \(\Delta S_{\text {univ }}\) for the process in vessel 2. (Hint: Think about the effect that a salt has on the freezing point of a solvent.)

Is \(\Delta S_{\text {surr favorable or unfavorable for exothermic reactions? }}\) Endothermic reactions? Explain.

The standard free energies of formation and the standard enthalpies of formation at \(298 \mathrm{K}\) for diffuoroacetylene \(\left(\mathrm{C}_{2} \mathrm{F}_{2}\right)\) and hexafluorobenzene \(\left(\mathrm{C}_{6} \mathrm{F}_{6}\right)\) are $$\begin{array}{ccc} & \Delta G_{f}^{\circ}(\mathrm{kJ} / \mathrm{mol}) & \Delta H_{f}^{\circ}(\mathrm{kJ} / \mathrm{mol}) \\ \hline \mathrm{C}_{2} \mathrm{F}_{2}(g) & 191.2 & 241.3 \\\ \mathrm{C}_{6} \mathrm{F}_{6}(g) & 78.2 & 132.8 \end{array}$$ For the following reaction: $$\mathrm{C}_{6} \mathrm{F}_{6}(g) \rightleftharpoons 3 \mathrm{C}_{2} \mathrm{F}_{2}(g)$$ a. calculate \(\Delta S^{\circ}\) at \(298 \mathrm{K}\). b. calculate \(K\) at 298 K. c. estimate \(K\) at \(3000 .\) K, assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

For each of the following pairs of substances, which substance has the greater value of \(S^{\circ} ?\) a. \(\mathrm{C}_{\text {graphite }}(s)\) or \(\mathrm{C}_{\text {diamond }}(s)\) b. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\) c. \(\mathrm{CO}_{2}(s)\) or \(\mathrm{CO}_{2}(g)\)

Liquid water at \(25^{\circ} \mathrm{C}\) is introduced into an evacuated, insulated vessel. Identify the signs of the following thermodynamic functions for the process that occurs: \(\Delta H, \Delta S, \Delta T_{\text {water }}, \Delta S_{\text {surr }}, \Delta S_{\text {univ. }}\)
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