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Chapter 17: Problem 31

Choose the substance with the larger positional probability in each case. a. 1 mole of \(\mathrm{H}_{2}\) (at STP) or 1 mole of \(\mathrm{H}_{2}\) (at \(\left.100^{\circ} \mathrm{C}, 0.5 \mathrm{~atm}\right)\) b. 1 mole of \(\mathrm{N}_{2}\) (at STP) or 1 mole of \(\mathrm{N}_{2}\) (at \(\left.100 \mathrm{~K}, 2.0 \mathrm{~atm}\right)\) c. 1 mole of \(\mathrm{H}_{2} \mathrm{O}(s)\) (at \(\left.0^{\circ} \mathrm{C}\right)\) or 1 mole of \(\mathrm{H}_{2} \mathrm{O}(l)\) (at \(\left.20^{\circ} \mathrm{C}\right)\)

Short Answer

Expert verified
a. 1 mole of H2 at 100°C and 0.5 atm has a larger positional probability. b. 1 mole of N2 at STP has a larger positional probability. c. 1 mole of H2O(l) at 20°C has a larger positional probability.

Step by step solution

01

(Case a: 1 mole of H2 at STP vs. 1 mole of H2 at 100°C and 0.5 atm)

The Ideal Gas Law can be applied to both states of H2. First, we must convert the temperature from Celsius to Kelvin: \(100°C + 273.15 = 373.15 K\). Now we can calculate the volume for each state: 1 mole of H2 at STP: \(V = \frac{nRT}{P} = \frac{(1)(0.0821)(273.15)}{1} = 22.42 L\) 1 mole of H2 at 100°C and 0.5 atm: \(V = \frac{nRT}{P}= \frac{(1)(0.0821)(373.15)}{0.5} = 61.31 L\) In this case, 1 mole of H2 at 100°C and 0.5 atm has a larger positional probability due to its larger volume at the given conditions.
02

(Case b: 1 mole of N2 at STP vs. 1 mole of N2 at 100 K and 2.0 atm)

Again, we will use the Ideal Gas Law to calculate the volume of each state: 1 mole of N2 at STP: \(V = \frac{nRT}{P} = \frac{(1)(0.0821)(273.15)}{1} = 22.42 L\) 1 mole of N2 at 100 K and 2.0 atm: \(V = \frac{nRT}{P} = \frac{(1)(0.0821)(100)}{2.0} = 4.11 L\) In this case, 1 mole of N2 at STP has a larger positional probability due to its larger volume at the given conditions.
03

(Case c: 1 mole of H2O(s) at 0°C vs. 1 mole of H2O(l) at 20°C)

For this case, we are comparing 1 mole of water in the solid (ice) state to the liquid state (water). Comparing the positional probabilities in different phases, we know that gases typically have the highest positional probability, followed by liquids and then solids. Between a solid and a liquid, the substance in the liquid state will typically have a larger positional probability because the molecules have greater mobility and are less confined in their positions. Therefore, in this case, 1 mole of H2O(l) at 20°C has a larger positional probability than 1 mole of H2O(s) at 0°C.

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Most popular questions from this chapter

Consider the reaction: $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) $$ At \(25^{\circ} \mathrm{C}, \Delta H^{\circ}=-92.50 \mathrm{~kJ}\). Which of the following statements is(are) true? a. This is an endothermic reaction. b. \(\Delta S^{\circ}\) for this reaction is negative. c. If the temperature is increased, the ratio \(\frac{\mathrm{PCl}_{5}}{\mathrm{PCl}_{3}}\) will increase. d. \(\Delta G^{\circ}\) for this reaction has to be negative at all temperatures. e. When \(\Delta G^{\circ}\) for this reaction is negative, then \(K_{\mathrm{p}}\) is greater than \(1.00\).

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) b. \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{HCl}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

The standard free energies of formation and the standard enthalpies of formation at \(298 \mathrm{~K}\) for difluoroacetylene \(\left(\mathrm{C}_{2} \mathrm{~F}_{2}\right)\) and hexafluorobenzene \(\left(\mathrm{C}_{6} \mathrm{~F}_{6}\right)\) are $$ \begin{array}{|lcc|} & \left.\Delta G_{\mathrm{f}}^{\circ}(\mathrm{k}] / \mathrm{mol}\right) & \Delta H_{\mathrm{f}}^{\circ}(\mathrm{kJ} / \mathrm{mol}) \\ \mathrm{C}_{2} \mathrm{~F}_{2}(g) & 191.2 & 241.3 \\ \mathrm{C}_{6} \mathrm{~F}_{6}(g) & 78.2 & 132.8 \\ \hline \end{array} $$ 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 \(298 \mathrm{~K}\). c. estimate \(K\) at \(3000 . \mathrm{K}\), assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

For mercury, the enthalpy of vaporization is \(58.51 \mathrm{~kJ} / \mathrm{mol}\) and the entropy of vaporization is \(92.92 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). What is the normal boiling point of mercury?

Consider the reaction: $$ \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{~S}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ for which \(\Delta H\) is \(-233 \mathrm{~kJ}\) and \(\Delta S\) is \(-424 \mathrm{~J} / \mathrm{K}\). a. Calculate the free energy change for the reaction \((\Delta G)\) at \(393 \mathrm{~K}\). b. Assuming \(\Delta H\) and \(\Delta S\) do not depend on temperature, at what temperatures is this reaction spontaneous?
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