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Chapter 17: Problem 94

The sublimation of carbon dioxide at \(-78^{\circ} \mathrm{C}\) is $$ \mathrm{CO}_{2}(s) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta H_{\mathrm{sub}}=62.4 \mathrm{~kJ} / \mathrm{mol} $$ Calculate \(\Delta S_{\text {sub }}\) when \(84.8 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) sublimes at this temperature.

Short Answer

Expert verified
The entropy change, \(\Delta S_{sub}\), during the sublimation of 84.8 g of carbon dioxide at \(-78^{\circ}C\) is calculated by utilising the thermodynamic equation relating \(\Delta H\), \(\Delta S\) and \(T\). The result will be in J/mol K.

Step by step solution

01

Analyze Given Information

Initially, we have the data \(\Delta H_{\text {sub }} = 62.4 \mathrm{~kJ} / \mathrm{mol}\), mass of \(CO_2 = 84.8 \mathrm{~g}\), and sublimation temperature \(T = -78^{\circ} \mathrm{C}\) that needs to be converted to Kelvin by adding 273 to the given Celsius temperature.
02

Calculate The Number Of Moles

Next, we need to calculate the number of moles (\(n\)) of carbon dioxide. This is done using the molar mass of carbon dioxide (44.01 g/mol) and the given amount of carbon dioxide (84.8 g). \[ n = \frac{mass}{Molar\ mass} = \frac{84.8 \ g}{44.01 \ g/mol} = 1.927 \ mol \]
03

Calculate Entropy Change

With the number of moles and temperature calculated, we can now find \(\Delta S_{sub}\). By definition, the entropy of sublimation is given by \[ \Delta S_{sub} = \frac{\Delta H_{sub}}{T} \] To find \(\Delta S_{sub}\), it's necessary to ensure that the units of \(\Delta H_{sub}\) and \(T\) are appropriate. Since \(\Delta H_{sub}\) is given in kJ/mol, we must convert it to J/mol as the unit for entropy is J/mol K. Similarly, make sure \(T\) is in Kelvin(K). After ensuring the correct units, plug the values into the formula: \[ \Delta S_{sub} = \frac{62.4 \times 10^3 \ J/mol}{(-78°C + 273) \ K} \]

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Most popular questions from this chapter

(a) Calculate \(\Delta G^{\circ}\) and \(K_{P}\) for the following equilibrium reaction at \(25^{\circ} \mathrm{C}\). The \(\Delta G_{\mathrm{f}}^{\circ}\) values are 0 for \(\mathrm{Cl}_{2}(g),-286 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{PCl}_{3}(g),\) and \(-325 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{PCl}_{5}(g)\) $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ (b) Calculate \(\Delta G\) for the reaction if the partial pressures of the initial mixture are \(P_{\mathrm{PCl}_{5}}=0.0029 \mathrm{~atm}\) \(P_{\mathrm{PCl}_{3}}=0.27 \mathrm{~atm},\) and \(P_{\mathrm{Cl}_{2}}=0.40 \mathrm{~atm}\)

From the values of \(\Delta H\) and \(\Delta S,\) predict which of the following reactions would be spontaneous at \(25^{\circ} \mathrm{C}\) : reaction \(\mathrm{A}: \Delta H=10.5 \mathrm{~kJ} / \mathrm{mol}, \Delta S=30 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} ;\) reaction \(\mathrm{B}: \Delta H=1.8 \mathrm{~kJ} / \mathrm{mol}, \Delta S=-113 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) If either of the reactions is nonspontaneous at \(25^{\circ} \mathrm{C}\), at what temperature might it become spontaneous?

(a) Over the years there have been numerous claims about "perpetual motion machines," machines that will produce useful work with no input of energy. Explain why the first law of thermodynamics prohibits the possibility of such a machine existing. (b) Another kind of machine, sometimes called a "perpetual motion of the second kind," operates as follows. Suppose an ocean liner sails by scooping up water from the ocean and then extracting heat from the water, converting the heat to electric power to run the ship, and dumping the water back into the ocean. This process does not violate the first law of thermodynamics, for no energy is created - energy from the ocean is just converted to electrical energy. Show that the second law of thermodynamics prohibits the existence of such a machine.

The \(K_{P}\) for the reaction $$ \mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} $$ is \(2.4 \times 10^{-3}\) at \(720^{\circ} \mathrm{C}\). What is the minimum partial pressure of \(\mathrm{N}_{2}\) required for the reaction to be spontaneous in the forward direction if the partial pressures of \(\mathrm{H}_{2}\) and \(\mathrm{NH}_{3}\) are \(1.52 \mathrm{~atm}\) and \(2.1 \times 10^{-2} \mathrm{~atm}\) respectively?

Derive the equation $$ \Delta G=R T \ln (Q / K) $$ where \(Q\) is the reaction quotient and describe how you would use it to predict the spontaneity of a reaction.
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