Name _________________________________________ Block _____________ Date ____________________
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
P1V1 = P2V2
V = (.98 atm x 10.3 L)/1.88 atm = 5.4 L
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
RT
D = (46 g/mol x .970 atm)/(.082 x 305 K) = 1.8 g/L
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)
(10)

+ 4H2(g)
(0)
b. 2KOH(s)
2(-425)
+ CO2(g)
(-393.5)
c. SO2(g) + 2H2S(g)
(-297)
2(-21)
d. Fe2O3(s)
(-826)

+ 6HCl(g)
6(-92)
12. From the heats of reaction:
x3(2H2 + O2 
N2(g) + 4H2O(l)
(0)
4(-286)

K2CO3(s) + H2O(g)
(-1150)
(-242)
ΔH = -148.5 kJ
(3/8) S8(s) + 2H2O(g)
3/ (102)
2(-242)
8
 2FeCl3(s)
2(-400)
ΔH = -106.75 kJ
+ 3H2O(g)
3(-242)
2H2O
ΔH = -148 kJ
H = -483.6 kJ)
H = +284.6 kJ)
 2O3
rev(3O2
ΔH = -1154 kJ
Calculate the heat of the following reaction
3H2
+ O3  3H2O
Reduce final equation (divide by 2)
13. Given the data:
x2(N2
+ O2
Rev(2NO
2N2O

2NO
+ O2  2NO2
 2N2 + O2
ΔH = -867.7 kJ
H = +180.7 kJ)
H = -113.1kJ)
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