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Chapter 5: Problem 91

At \(20^{\circ} \mathrm{C}\) (approximately room temperature) the average velocity of \(\mathrm{N}_{2}\) molecules in air is \(1050 \mathrm{mph}\). (a) What is the average speed in \(\mathrm{m} / \mathrm{s}\) ? (b) What is the kinetic energy (in J) of an \(\mathrm{N}_{2}\) molecule moving at this speed? (c) What is the total kinetic energy of \(1 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) molecules moving at this speed?

Short Answer

Expert verified
(a) The average speed in meters per second is \(1050 \times \frac{1609.34}{3600} \approx 469.62 \, \mathrm{m/s}\). (b) The kinetic energy of a single N2 molecule is approximately \(KE \approx 6.389 \times 10^{-21} \, \mathrm{J}\). (c) The total kinetic energy of one mole of N2 molecules moving at this speed is approximately \(8.452 \times 10^{3} \, \mathrm{J}\).

Step by step solution

01

Convert velocity from mph to m/s

To convert the velocity given in mph (miles per hour) to m/s (meters per second), you can use the following conversion factor: 1 mile = 1609.34 meters and 1 hour = 3600 seconds. The average velocity of an N2 molecule is 1050 mph. Let's now convert it to m/s: Average velocity (m/s) = 1050 mph * (1609.34 m / 1 mile) * (1 hour / 3600 s) Calculating this will give us the average speed in m/s.
02

Calculate kinetic energy

Now, let's find the kinetic energy of a single N2 molecule moving at this speed. The formula for the kinetic energy is: KE = \(\frac{1}{2}\) * m * v^2 where KE is the kinetic energy, m is the mass of the molecule, and v is the average velocity. We are given the average velocity of N2 molecules in m/s from part (a) and the mass of one N2 molecule can be calculated as: Mass of one N2 molecule = (28 g/mol) * (1 mol / 6.022 * 10^23 molecules) * (1 kg / 1000 g) Now, substitute the values into the equation for the kinetic energy and calculate the result.
03

Calculate the total kinetic energy of one mole

For part (c), we need to find the total kinetic energy of one mole of N2 molecules moving at the average velocity given. We know there is exactly 6.022 × 10^23 molecules per mole. So, the total kinetic energy for 1 mole of N2 molecules can be found by multiplying the kinetic energy of each single N2 molecule (found in part (b)) by the number of molecules in one mole: Total Kinetic Energy = Kinetic Energy of 1 molecule * number of molecules in 1 mole Now, just plug in the values and calculate the result.

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Most popular questions from this chapter

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat \(\mathrm{KClO}_{3}\) : $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{~kJ} $$ For this reaction, calculate \(\Delta H\) for the formation of (a) \(1.36\) \(\mathrm{mol}\) of \(\mathrm{O}_{2}\) and (b) \(10.4 \mathrm{~g}\) of \(\mathrm{KCl}\). (c) The decomposition of \(\mathrm{KClO}_{3}\) proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of \(\mathrm{KClO}_{3}\) from \(\mathrm{KCl}\) and \(\mathrm{O}_{2}\), is likely to be feasible under ordinary conditions? Explain your answer.

(a) When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter (Figure 5.18), the temperature rises \(1.642^{\circ} \mathrm{C}\). When a \(0.265-\mathrm{g}\) sample of caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{~N}_{4}\), is burned, the temperature rises \(1.525^{\circ} \mathrm{C}\). Using the value \(26.38 \mathrm{~kJ} / \mathrm{g}\) for the heat of combustion of benzoic acid, calculate the heat of combustion per mole of caffeine at constant volume. (b) Assuming that there is an uncertainty of \(0.002^{\circ} \mathrm{C}\) in each temperature reading and that the masses of samples are measured to \(0.001 \mathrm{~g}\), what is the estimated uncertainty in the value calculated for the heat of combustion per mole of caffeine?

Consider the following reaction: $$ 2 \mathrm{CH}_{3} \mathrm{OH}(g) \longrightarrow 2 \mathrm{CH}_{4}(g)+\mathrm{O}_{2}(g) \quad \Delta H=+252.8 \mathrm{~kJ} $$ (a) Is this reaction exothermic or endothermic? (b) Calculate the amount of heat transferred when \(24.0 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{OH}(g)\) is decomposed by this reaction at constant pressure. (c) For a given sample of \(\mathrm{CH}_{3} \mathrm{OH}\), the enthalpy change during the reaction is \(82.1 \mathrm{~kJ}\). How many grams of methane gas are produced? (d) How many kilojoules of heat are released when \(38.5 \mathrm{~g}\) of \(\mathrm{CH}_{4}(g)\) reacts completely with \(\mathrm{O}_{2}(g)\) to form \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) at constant pressure? 5.45 When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates: $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \operatorname{AgCl}(s) \quad \Delta H=-65.5 \mathrm{~kJ} $$ (a) Calculate \(\Delta H\) for the production of \(0.450 \mathrm{~mol}\) of \(\mathrm{AgCl}\) by this reaction. (b) Calculate \(\Delta H\) for the production of \(9.00 \mathrm{~g}\) of \(\mathrm{AgCl}\). (c) Calculate \(\Delta H\) when \(9.25 \times 10^{-4} \mathrm{~mol}\) of \(\mathrm{AgCl}\) dissolves in water.

Using values from Appendix C, calculate the value of \(\Delta H^{\circ}\) for each of the following reactions: (a) \(\mathrm{CaO}(s)+2 \mathrm{HCl}(g) \longrightarrow \mathrm{CaCl}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(4 \mathrm{FeO}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) (c) \(2 \mathrm{CuO}(s)+\mathrm{NO}(g) \longrightarrow \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{NO}_{2}(g)\) (d) \(4 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

How much work (in J) is involved in a chemical reaction if the volume decreases from \(5.00\) to \(1.26 \mathrm{~L}\) against a constant pressure of \(0.857 \mathrm{~atm}\) ?
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