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Chapter 19: Problem 25

The normal boiling point of \(\mathrm{Br}_{2}(l)\) is \(58.8{ }^{\circ} \mathrm{C},\) and its molar enthalpy of vaporization is \(\Delta H_{\text {vap }}=29.6 \mathrm{~kJ} /\) mol. (a) When \(\mathrm{Br}_{2}(l)\) boils at its normal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when \(1.00 \mathrm{~mol}\) of \(\mathrm{Br}_{2}(l)\) is vaporized at \(58.8{ }^{\circ} \mathrm{C}\).

Short Answer

Expert verified
(a) The entropy increases during boiling of Br2 at its normal boiling point because ΔS = ΔH / T, and both ΔH and T have positive values. (b) To calculate the value of ΔS when 1.00 mol of Br2 is vaporized at 58.8 °C, first convert the temperature to Kelvin and then use the equation ΔS = ΔH / T: T(K) = 58.8 + 273.15 = 331.95 K ΔS = (29.6 kJ/mol) / (331.95 K) = (29.6 × 10^3 J/mol) / (331.95 K) ≈ 89.16 J/(mol·K) The entropy change when 1 mol of Br2 is vaporized at 58.8 °C is 89.16 J/(mol·K).

Step by step solution

01

(Step 1: Determine if the entropy increases or decreases during boiling)

(At the normal boiling point, we know the change in Gibbs free energy (ΔG) is zero because the process occurs at equilibrium. Using the equation ΔG = ΔH - TΔS, and solving for ΔS, we get: ΔS = ΔH / T Since ΔH (enthalpy of vaporization) has a positive value and T (temperature) is also positive because it is in Kelvin, we can conclude that ΔS has a positive value. Thus, the entropy increases during boiling.)
02

(Step 2: Calculate the value of ΔS for vaporization of 1 mole of Br2 at 58.8 °C)

(First, we need to convert the temperature from Celsius to Kelvin: T(K) = 58.8 + 273.15 = 331.95 K Next, we use the equation ΔS = ΔH / T, where ΔH is the enthalpy of vaporization (29.6 kJ/mol) and T(K) is the temperature in Kelvin: ΔS = (29.6 kJ/mol) / (331.95 K) Now, we convert ΔH units from kJ/mol to J/mol to obtain ΔS in J/(mol·K): ΔS = (29.6 × 10^3 J/mol) / (331.95 K) Finally, we calculate the value of ΔS: ΔS = 89.16 J/(mol·K) So, the entropy change when 1 mol of Br2 is vaporized at 58.8 °C is 89.16 J/(mol·K).)

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Most popular questions from this chapter

(a) What sign for \(\Delta S\) do you expect when the pressure on 0.600 mol of an ideal gas at \(350 \mathrm{~K}\) is increased isothermally from an initial pressure of 0.750 atm? (b) If the final pressure on the gas is 1.20 atm, calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change? Explain.

Using the data in Appendix \(C\) and given the pressures listed, calculate \(\Delta G^{\circ}\) for each of the following reactions: $$ \begin{array}{l} \text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \\ \quad P_{\mathrm{N}_{2}}=2.6 \mathrm{~atm}, P_{\mathrm{H}_{2}}=5.9 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=1.2 \mathrm{~atm} \\ \text { (b) } 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) \\ \quad P_{\mathrm{N}_{2} \mathrm{H}_{4}}=P_{\mathrm{NO}_{2}}=5.0 \times 10^{-2} \mathrm{~atm} \\ \quad P_{\mathrm{N}_{2}}=0.5 \mathrm{~atm}, P_{\mathrm{H}_{2} \mathrm{O}}=0.3 \mathrm{~atm} \\ \text { (c) } \mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \\ \quad P_{\mathrm{N}_{2} \mathrm{H}_{4}}=0.5 \mathrm{~atm}, P_{\mathrm{N}_{2}}=1.5 \mathrm{~atm}, P_{\mathrm{H}_{2}}=2.5 \mathrm{~atm} \end{array} $$

(a) What is the meaning of the standard free-energy change, \(\Delta G^{\circ},\) as compared with \(\Delta G\) ? (b) For any process that occurs at constant temperature and pressure, what is the significance of \(\Delta G=0 ?(c)\) For a certain process, \(\Delta G\) is large and negative. Does this mean that the process necessarily occurs rapidly?

Using data in Appendix C, calculate \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for each of the following reactions. In each case show that \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} .\) (a) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)\) (b) \(\mathrm{C}(s,\) graphite \()+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)\) (c) \(2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{POCl}_{3}(g)\) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

Using \(S^{\circ}\) values from Appendix C, calculate \(\Delta S^{\circ}\) values for the following reactions. In each case account for the sign of \(\Delta S^{\circ} .\) (a) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)\) (b) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) (c) \(\mathrm{Be}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{BeO}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\)
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