TTh
MW
MW
am mid pm
Student Name _______________________________ _______________
Lab Section (circle one)
Chapter 8 Homework Problems
CHEM 1A
1. Consider the following balanced thermochemical equation:
Fe2O3 (s) + 3 CO (g)  2 Fe (s) + 3 CO2 (g)
HRxn = – 24.8 kJ
(a) Is heat absorbed or released in the reaction (circle one)?
(b) If the heat was written into the equation, would it appear on the reactant side or the
product side of the equation (circle one)?
(c) What is HRxn for the reverse reaction? _____________
(d) What is H when 185.9 g of CO reacts with excess Fe2O3?
2. The reaction depicted to the right is exothermic:
(a) Write a balanced equation for the
reaction (red spheres represent A atoms
and ivory spheres represent B atoms).
(b) Label the sign (+ or –) of each variable in the equation below. If G depends on the
temperature, enter a question mark (?).
G = H – T S
( ) = ( ) – ( )( )
(c) Will the reaction be spontaneous at all temperatures, at low temperatures only, at high
temperatures only, or at no temperatures (circle one)?
3. A reaction is exothermic, yet non-spontaneous under the current conditions.
(a) What must the sign of S be? Fill in: S ( )
(b) Are the current conditions likely to be at high temperatures or low temperatures (circle
one)?
CHEM 1A – Chapter 8 – Page 1 of 6
4. Circle one for each of the following:
(a) Is the heat capacity of a substance an intensive or an extensive property?
(b) Is the specific heat of a substance an intensive or an extensive property?
5. A 30.5 g sample of an alloy at 93.0°C is placed into 50.0 g of water at 22.0°C in an insulated coffee
cup with a heat capacity of 9.2 J/°C. If the final temperature of the system is 31.1°C, what is the
specific heat capacity of the alloy?
6. A 45.2 g sample of iron and a 68.2 g sample of aluminum, both at 98.0°C, are placed into 150.0 g of
water at 23.2°C in an insulated coffee cup with a heat capacity of 8.7 J/°C. Calculate the final
temperature of the system. CFe = 0.449 J/g·°C, CAl = 0.897 J/g·°C
CHEM 1A – Chapter 8 – Page 2 of 6
7. When 25.0 mL of 0.500 M H2SO4 is added to 25.0 mL of 1.00 M KOH (stoichiometric amounts) in a
coffee cup calorimeter at 23.50°C, the temperature rises to 30.17°C.
(a) Write a balanced chemical reaction for the acid-base neutralization occurring in the
calorimeter.
(b) Given the following assumptions, calculate HRxn in kJ/mol H2SO4.
Assume: all heat is transferred to the water, the solution volumes are additive, and the
resulting solution has the same density and specific heat capacity as pure water
Hint: HRxn =
– qw
# mol H2SO4
8. Calculate the change in internal energy (in kJ) when 89.5 g C2H4 react with excess HCl at 1.005 atm
of pressure with a volume change of – 71.5 L.
C2H4 (g) + HCl (g)  C2H5Cl (g)
CHEM 1A – Chapter 8 – Page 3 of 6
H = – 72.3 kJ
9. Calculate HRxn for
2 NOCl (g)  N2 (g) + O2 (g) + Cl2 (g)
given the following set of reactions:
½ N2 (g) + ½ O2 (g)  NO (g)
NO (g) + ½ Cl2 (g)  NOCl (g)
H = 90.3 kJ
H = – 38.6 kJ
10. Use Hess’s law to calculate the standard heat of formation for aqueous nitric acid.
(a) Write the balanced equation for the formation of 1 mole of nitric acid from elements in their
natural state (you may need to use fractional coefficients for the reactants).
(b) Complete and use the given reactions below:
3 NO2 (g) + H2O (l)  2 HNO3 (aq) + NO (g)
2 NO (g) + O2 (g)  2 NO2 (g)
4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (l)
 NH3 (g)
 H2O (l)
CHEM 1A – Chapter 8 – Page 4 of 6
H = – 137.3 kJ
H = – 116.2 kJ
H = – 1165.2 kJ
H = – 46.1 kJ
H = – 285.8 kJ
11. Calculate H
for the reaction below using H data from Table B1 of Appendix B in the back of
your textbook.
SiO2 (s) + 4 HF (g)  SiF4 (g) + 2 H2O (l)
12. The following equation is for the combustion of 1 mole of acetylene (C2H2) in air:
C2H2 (g) + 5/2 O2 (g)  2 CO2 (g) + H2O (g)
Calculate H of C2H2 (g), given H of CO2 (g) = – 393.5 kJ/mol
and H of H2O (g) = – 241.8 kJ/mol
CHEM 1A – Chapter 8 – Page 5 of 6
H
= – 1255.8 kJ
13. Although it is noble gas and inert under normal conditions, xenon can form compounds with highly
electronegative elements like oxygen and fluorine. XeF6 is formed by the direct reaction of the
elements.
(a) Write the balanced equation between xenon and fluorine to form 1 mole of XeF6.
(b) Calculate the Xe–F bond energy in XeF6, given that the heat of formation is – 402 kJ/mol,
and the F–F bond energy is 159 kJ/mol.
14. Isopropyl (rubbing) alcohol, is prepared industrially by the
process shown to the below. Calculate H using the average
bond dissociation energies given in table to the right.
Bond
C≡C
C=C
C–C
C–H
C=O
C–O
O–H
O=O
CHEM 1A – Chapter 8 – Page 6 of 6
Bond Energy
(kJ/mol)
839
614
347
413
745
799 (in CO2)
358
467
498