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Chapter 19: Problem 88

For each of the following processes, indicate whether the signs of \(\Delta S\) and \(\Delta H\) are expected to be positive, negative, or about zero. (a) A solid sublimes. (b) The temperature of a sample of \(\mathrm{Co}(s)\) is lowered from \(60^{\circ} \mathrm{C}\) to $25^{\circ} \mathrm{C} .$ (c) Ethyl alcohol evaporates from a beaker. (d) A diatomic molecule dissociates into atoms. (e) A piece of charcoal is combusted to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\).

Short Answer

Expert verified
a) ΔS > 0, ΔH > 0 b) ΔS < 0, ΔH < 0 c) ΔS > 0, ΔH > 0 d) ΔS > 0, ΔH > 0 e) ΔS > 0, ΔH < 0

Step by step solution

01

a) A solid sublimes.

In sublimation, a solid changes directly into a gas without passing through a liquid state. This results in an increase in the disorder of the system, so the entropy (∆S) will be positive. Since this process is endothermic, meaning it requires energy input to break the bonds and change the solid into a gas, the enthalpy (∆H) is positive as well. Result: ΔS > 0, ΔH > 0
02

b) The temperature of a sample of Co(s) is lowered from 60°C to 25°C

In this process, the temperature of the sample of cobalt(s) is decreasing. As the temperature decreases, the motion of the particles within the solid slows down. This results in a decrease in disorder, so the entropy (∆S) will be negative. The cooling process releases heat, making it an exothermic process, so the enthalpy (∆H) is negative as well. Result: ΔS < 0, ΔH < 0
03

c) Ethyl alcohol evaporates from a beaker

When ethyl alcohol evaporates, its molecules go from a liquid state to a gaseous state. Gas molecules have more freedom of motion and are more disordered compared to liquid molecules. Therefore, the entropy (∆S) will be positive. This process is endothermic because it requires input of heat energy to break intermolecular forces, so the enthalpy (∆H) is positive as well. Result: ΔS > 0, ΔH > 0
04

d) A diatomic molecule dissociates into atoms

When a diatomic molecule dissociates into individual atoms, the number of particles in the system increases and the particles have more freedom of motion. This results in an increase in disorder, so the entropy (∆S) will be positive. As bonds must be broken in order to separate the atoms, this process requires energy input, so the enthalpy (∆H) is also positive. Result: ΔS > 0, ΔH > 0
05

e) A piece of charcoal is combusted to form CO2(g) and H2O(g)

Combustion of charcoal is an exothermic reaction, meaning it releases heat. Therefore, the enthalpy (∆H) will be negative. In the combustion process, the complex carbon structure of charcoal is broken down into smaller, simpler molecules (CO2 and H2O gases). This results in an increase in disorder, so the entropy (∆S) will be positive. Result: ΔS > 0, ΔH < 0

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Most popular questions from this chapter

Indicate whether \(\Delta G\) increases, decreases, or does not change when the partial pressure of \(\mathrm{H}_{2}\) is increased in each of the following reactions: (a) $\mathrm{H}_{2}(g)+\mathrm{NiO}(s) \longrightarrow \mathrm{Ni}(s)+\mathrm{H}_{2} \mathrm{O}(g)$ (b) $\mathrm{H}_{2}(g)+\mathrm{S}(s) \longrightarrow \mathrm{H}_{2} \mathrm{~S}(g)$ (c) $\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g)$

Using data from Appendix \(\mathrm{C}\), write the equilibrium-constant expression and calculate the value of the equilibrium constant and the free- energy change for these reactions at \(298 \mathrm{~K}:\) (a) $\mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g)$ (b) $2 \mathrm{HBr}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\mathrm{Br}_{2}(g)$ (c) $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$

The reaction $2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)$ is highly spontaneous. A classmate calculates the entropy change for this reaction and obtains a large negative value for \(\Delta S^{\circ}\). Did your classmate make a mistake in the calculation? Explain.

Consider a system that consists of two standard playing dice, with the state of the system defined by the sum of the values shown on the top faces. (a) The two arrangements of top faces shown here can be viewed as two possible microstates of the system. Explain. (b) To which state does each microstate correspond? (c) How many possible states are there for the system? (d) Which state or states have the highest entropy? Explain. (e) Which state or states have the lowest entropy? Explain. (f) Calculate the absolute entropy of the two-dice system.

For each of the following pairs, predict which substance has the higher entropy per mole at a given temperature: (a) \(\mathrm{I}_{2}(s)\) or \(\mathrm{I}_{2}(g)\) (b) \(\mathrm{O}_{2}(g)\) at \(50.7 \mathrm{kPa}\) or \(\mathrm{O}_{2}\) at \(101.3 \mathrm{kPa}\) (c) 1 molof \(\mathrm{N}_{2}\) in 22.4 Lor \(1 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) in \(44.8 \mathrm{~L}\). (d) \(\mathrm{CH}_{3} \mathrm{OH}(I)\) or \(\mathrm{CH}_{3} \mathrm{OH}(s)\)
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