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Chapter 12: Problem 22

A vapor volume of 1.17 L forms when a sample of liquid acetonitrile, \(\mathrm{CH}_{3} \mathrm{CN},\) absorbs \(1.00 \mathrm{kJ}\) of heat at its normal boiling point \(\left(81.6^{\circ} \mathrm{C} \text { and } 1 \mathrm{atm}\right) .\) What is \(\Delta H_{\text {vap }}\) in kilojoules per mole of \(\mathrm{CH}_{3} \mathrm{CN} ?\)

Short Answer

Expert verified
The enthalpy of vaporization (\( \Delta H_{\text {vap }} \)) of acetonitrile is the result obtained by dividing the heat absorbed (1.00 kJ) by the number of moles of acetonitrile calculated in step 1.

Step by step solution

01

Calculate the number of moles

First, we need to calculate the number of moles of acetonitrile that produces 1.17L of vapor at 1 atm and 81.6°C. We use the ideal gas law here: \(PV = nRT\). Rearranging the equation, we get \(n=PV/RT\). Substituting the given values (keep in mind to use correct units - temperature should be in Kelvin, volume in liters, and remember R = 0.0821 L.atm/K.mol):\(n = (1 atm)(1.17 L) / ((0.0821 L.atm/K.mol)(81.6°C + 273.15))\)This gives the number of moles of acetonitrile that forms 1.17L vapor.
02

Calculate enthalpy of vaporization

Now we've obtained that the number of moles of acetonitrile is \(n\). We know that 1.00 kJ of heat is absorbed for this many moles to vaporize. Therefore,\(\Delta H_{\text {vap }} = Q/n\)Substitute Q=1.00 kJ and \(n\) from the previous step. This will give us the enthalpy of vaporization of acetonitrile.

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Most popular questions from this chapter

If the triple point pressure of a substance is greater than 1 atm, which two of the following conclusions are valid? (a) The solid and liquid states of the substance cannot coexist at equilibrium. (b) The melting point and boiling point of the substance are identical. (c) The liquid state of the substance cannot exist. (d) The liquid state cannot be maintained in a beaker open to air at 1 atm pressure. (e) The melting point of the solid must be greater than \(0^{\circ} \mathrm{C}\) (f) The gaseous state at 1 atm pressure cannot be condensed to the solid at the triple point temperature.

The following data are given for \(\mathrm{CCl}_{4}\). Normal melting point, \(-23^{\circ} \mathrm{C} ;\) normal boiling point, \(77^{\circ} \mathrm{C} ;\) density of liquid \(1.59 \mathrm{g} / \mathrm{mL} ; \Delta H_{\text {fus }}=3.28 \mathrm{kJ} \mathrm{mol}^{-1} ;\) vapor pressure at \(25^{\circ} \mathrm{C}, 110\) Torr. (a) What phases-solid, liquid, and/or gas-are present if \(3.50 \mathrm{g} \mathrm{CCl}_{4}\) is placed in a closed \(8.21 \mathrm{L}\) container at \(25^{\circ} \mathrm{C} ?\) (b) How much heat is required to vaporize 2.00 L of \(\mathrm{CCl}_{4}(\mathrm{l})\) at its normal boiling point?

A crystalline solid contains three types of ions, \(\mathrm{Na}^{+}, \mathrm{O}^{2-},\) and \(\mathrm{Cl}^{-}\). The solid is made up of cubic unit cells that have \(\mathrm{O}^{2-}\) ions at each corner, \(\mathrm{Na}^{+}\) ions at the center of each face, and \(\mathrm{Cl}^{-}\) ions at the center of the cells. What is the chemical formula of the compound? What are the coordination numbers for the \(\mathrm{O}^{2-}\) and \(\mathrm{Cl}^{-}\) ions? If the length of one edge of the unit cell is \(a,\) what is the shortest distance from the center of a \(\mathrm{Na}^{+}\) ion to the center of an \(\mathrm{O}^{2-}\) ion? Similarly, what is the shortest distance from the center of a \(\mathrm{Cl}^{-}\) ion to the center of an \(\mathrm{O}^{2-}\) ion?

Would you expect an ionic solid or a network covalent solid to have the higher melting point?

Briefly describe each of the following phenomena or methods: (a) capillary action; (b) polymorphism; (c) sublimation; (d) supercooling; (e) determining the freezing point of a liquid from a cooling curve.
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