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AP Chemistry Review Ch 5 & 6: Gas Laws & Thermochemistry

Chapter 5

1. A fixed quantity of gas at 23oC exhibits a pressure of 748 torr and occupies a volume of 10.3 L.

a. Use Boyle’s law to calculate the volume the gas will occupy at 23oC if the pressure is increased to

1.88 atm.

Boyle’s Law V = (.98 atm x 10.3 L)/1.88 atm = 5.4 L

P1V1 = P2V2

Don’t forget to convert Torr to Atm

b. Use Charles’s law to calculate the volume the gas will occupy if the temperature is increased to

165oC while the pressure is held constant.

Charles Law

V1 = V2 V = (438 K x 10.3 L)/296 K=15.2 L

T1 T2

Don’t forget to convert oC to K

2. The Hindenburg was a hydrogen-filled dirigible that exploded in 1937. If the Hindenburg held 2.0 x 105 m3

of hydrogen gas at 23oC and 1.0 atm, what mass of hydrogen was present?

PV=nRT n=(1 atm x 2x108 m3)/(.082 x 296K) = 8.2 x 106 mol H2

8.2 x 106 mol H2 x 2g H2 = 1.65 x 107 g H2

Don’t forget to convert m3 to dm3 (1 m3 = 1000 dm3) and oC to K

3. a. Calculate the density of NO2 gas at 0.970 atm and 32oC.

D = MP D = (46 g/mol x .970 atm)/(.082 x 305 K) = 1.8 g/L

RT

Don’t forget to calculate molar mass of NO2

b. Calculate the molar mass of a gas if 2.50 g occupies 0.875 L at 685 torr and 35oC

M = DRT M = (2.9 g/L x .082 x 308K)/.9 atm = 81.4 g/mol

P

Don’t forget to calculate density (d=m/V) , convert oC to K and torr to atm.

4. The metabolic oxidation of glucose in our bodies produces CO2, which is expelled from our lungs as a gas.

Calculate the volume of dry CO2 produced at body temperature (37oC) and 0.970 atm when 24.5 g of

glucose is consumed in the reaction.

You need a balanced equation:

C6H12O6 + 6O2 ( 6CO2 + 6H2O

Convert 24.5 g glucose to mol CO2 using stoichiometry:

24.5 g glucose x (1mol glucose/180 g glucose) x (6 mol CO2/1 mol glucose) = .82 mol CO2

Use PV=nRT to adjust for conditions given in problem:

V = (.82 mol x .082 x 310K)/(.97 atm) = 21.5 atm

5. Hydrogen is produced when zinc reacts with sulfuric acid. If 159 mL of wet H2 is collected over water at

24oC and a pressure of 738 torr, how many grams of Zn have been consumed?

You need a balanced equation:

Zn + H2SO4 ( ZnSO4 + H2

Use Ptotal = PH2O + PH2 to calculate pressure of H2 gas and then convert to atm:

738 torr = 22.38 torr + PH2

PH2 = 715.6 torr/760 torr = .94 atm

Use PV=nRT to calculate moles of H2:

n = (.94 atm x .159L)/(.082 x 297 K)= .006 mol H2

Since mol Zn = mol H2 based on balanced equation:

.006 mol Zn x 65.3 g Zn = .4 g Zn needed

6. At an underwater depth of 250 ft, the pressure is 8.38 atm. What should the mole percent of oxygen be in

the diving gas for the partial pressure of oxygen in the mixture to be 0.21 atm, the same as in air at 1 atm?

XO2 = PO2 = nO2

Ptotal ntotal

.21 atm / 8.38 atm = .025 = 2.5 % O2 in the mix

7. A mixture containing 0.538 mol He(g), 0.315 mol Ne(g) and 0.103 mol Ar(g) is confined in a 7.00 L vessel at

25oC.

a. Calculate the partial pressure of each of the gases in the mixture

PHe = (.538 mol x .082 x 298K)/7L = 1.88 atm PNe = (.315 mol x .082 x 298K)/7L = 1.1 atm

PAr = (.103 mol x .082 x 298K)/7L = .35 atm

b. Calculate the total pressure of the mixture

Ptotal = PHe + PAr + PNe = 1.88 atm + .35 atm + 1.1 atm = 3.3 atm

8. Vessel A contains CO(g) at 0oC and 1 atm. Vessel B contains SO2(g) at 20oC and 0.5 atm. The two vessels

have the same volume.

a. Which vessel contains more molecules?

Vessel A has more molecules. (Lower T and higher P)

b. Which contains more mass?

Vessel B has more mass (higher molar mass)

c. In which vessel is the average kinetic energy of the molecules higher?

Vessel B (20oC greater than 0oC)

d. In which vessel is the average speed of the molecules higher?

Vessel A has higher speed (UA/UB = 1.46)

Chapter 6

9. Consider the following reaction:

2Mg(s) + O2(g) ( 2MgO(s) (H = -1204 kJ

a. Is this reaction exothermic or endothermic?

Exothermic (ΔH is negative, heat was given off)

b. Calculate the amount of heat transferred when 2.4 g of Mg(s) reacts at constant pressure.

2.4 g Mg x (1 mol Mg/24g Mg)x(-1204 kJ/1 mol) = -60.2 kJ

c. How many grams of MgO are produced during an enthalpy change of -96.0kJ?

-96.0 kJ x (2 mol MgO/-1204 kJ) x (40 g MgO/1 mol MgO) = 6.4 g MgO

d. How many kilojoules of heat are absorbed when 7.5 g of MgO are decomposed into Mg and O2 at

constant pressure? This is the reverse of the above reaction….so reverse sign on ΔH

7.5 g MgO x (1 mol MgO/40 g MgO) x (+1204 kJ/2 mol MgO) = +112.8 kJ

10. Calculate (E and determine whether the process is endothermic or exothermic:

a. A system releases 113 kJ of heat into the surroundings and does 39 kJ of work on the surroundings.

ΔE = -113 kJ + (-39 kJ) = -152 kJ exothermic

b. q = 1.62 kJ and w = -874 J

ΔE = 1.62 kJ + -0.874 kJ = +0.746 kJ endothermic

c. The system absorbs 77.5 kJ of heat while doing 63.5 kJ of work on the surroundings

ΔE = 77.5 kJ + -63.5 kJ = + 14 kJ endothermic

11. Using values from Appendix 4, calculate the value of (Ho for each of the following reactions:

a. N2O4(g) + 4H2(g) ( N2(g) + 4H2O(l)

(10) (0) (0) 4(-286) ΔH = -1154 kJ

b. 2KOH(s) + CO2(g) ( K2CO3(s) + H2O(g)

2(-425) (-393.5) (-1150) (-242) ΔH = -148.5 kJ

c. SO2(g) + 2H2S(g) ( (3/8) S8(s) + 2H2O(g)

(-297) 2(-21) 3/8(102) 2(-242) ΔH = -106.75 kJ

d. Fe2O3(s) + 6HCl(g) ( 2FeCl3(s) + 3H2O(g)

(-826) 6(-92) 2(-400) 3(-242) ΔH = -148 kJ

12. From the heats of reaction:

x3(2H2 + O2 ( 2H2O (H = -483.6 kJ)

rev(3O2 ( 2O3 (H = +284.6 kJ)

Calculate the heat of the following reaction

3H2 + O3 ( 3H2O

Reduce final equation (divide by 2) ΔH = -867.7 kJ

13. Given the data:

x2(N2 + O2 ( 2NO (H = +180.7 kJ)

Rev(2NO + O2 ( 2NO2 (H = -113.1kJ)

2N2O ( 2N2 + O2 (H = -163.2 kJ

use Hess’s law to calculate (H for the reaction

N2O + NO2 ( 3NO

Reduce final equation (divide by 2) ΔH = +155.65 kJ

14. A 46.2 g sample of copper is heated to 95.4oC and then placed in a calorimeter containing 75.0 g water at

19.6oC. The final temperature of the water and metal is 21.8oC. Calculate the specific heat capacity of

copper, assuming that all the heat lost by the copper is gained by the water.

Heat lost by Cu = heat gained by H2O

qH2O = (75 g)(2.2oC)(4.18) = 689.7 kJ

Cp for Cu = 689.7 kJ / (46.2 g x 73.6 oC) = .203 J/g oC

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