Thermochemistry Study Questions and Problems - KEY
1.
Describe in sentence format the transfer of energy that takes place when:
a.
A saucepan of water is heated to boiling
the chemical energy of the burner transfers kinetic energy to the water molecules
causing an increase in temperature and eventually enough energy to break and escape
the intermolecular attractions between water molecules.
b.
A chemical reaction liberates heat. In this exothermic reaction the chemical
potential energy of the system is converted into thermal energy of the system and its
surroundings
2.
If the temperature of 50.0 gram block of aluminum increases by 10.9 K when heated by 500
joules, calculate the:
a.
heat capacity of the aluminum block
500J/10.9 K = 45.9 JK-1
b.
molar heat capacity of aluminum
45.9 JK-1 * 26.98gmol-1/50.0g=24.8 JK-1mol-1
c.
specific heat of aluminum
500J/(10.0K*50.0g)=0.917 JK-1g—1
3.
The specific heat of gold is 0.128 JK-1g-1 and the specific heat of iron is 0.451 JK-1g—1.
Calculate the molar heat capacities of these two metals and compare to the value for
aluminum calculated in question #2.
0.128 JK-1g—1*196.97g/mol=25.2 JK-1mol—1
0.451 JK-1g—1* 55.85 g/mol = 25.2 JK-1mol—1
The molar heat capacities of all metals at room temperature are approximately the
same (theoretically = 3R(gas law constant 8.314jK-1mol-1) = 24.9 JK-1mol—1,
this is known as the Law of Dulong and Petit.
4.
Calculate the heat necessary to change the temperature of one kg of iron from 25 C to 1000 C
(FYI: the melting point of Fe=1811 K). The specific heat of iron is 0.451 JK-1g—1.
Heat and temperature are related by the heat capacity, where the heat capacity equals
the specific heat capacity x mass…note: heat capacity does not have a mass unit!
Energy = 0.451 JK-1g—1 * 1000 g * (1000-25)K = 440,000 = 440 kJ
note: Kelvin and Celsius temperature changes are equivalent degree by degree)
5.
If a 40 g block of copper at 100 C is added to 100 grams of water at 25 C, calculate the final
temperature assuming no heat is lost to the surroundings. The specific heat of copper is 0.385
JK-1g—1 and the specific heat of water is 4.185 JK-1g—1.
0.385 JK-1g—1*40 g*(100-Tf) = 4.184 JK-1g—1* 100 g *(Tf- 25 C) = 27.7 C.
6.
Calculate the amount of heat necessary to melt 27.0 grams of ice if the heat of fusion of ice is
6.009 kJ/mol.
6.009kJmol-1* 1.50 mol = 9.014 kJ.
7.
If 27.0 grams of ice at 0 C is added to 123 grams of water at 100 C in an insulated container,
calculate the final temperature. Assume that the specific heat of water is 4.185 JK-1g—1.
(Hint: utilize the information from problem #6.)
Use energy from the melting of ice from #6 = 9.014 kJ + temperature change calc.
Heat required (absorbed) to heat the water produced = 4.184 JK-1g—1 * 27.0 g * (T-O) C
Heat required (lost) by the hot water = 4.184 JK-1g—1 * 123 g * (100 – T) C
9014 + 113T = 51463 - 515T T=67.6 C
8.
A 50 g block of an unknown metal alloy at 100 C is dropped into an insulated flask containing
approximately 200 g of ice. It was determined that 10.5 g of the ice melted. What is the
specific heat capacity of the unknown alloy? (Hint: use info from Problem #6.)
50g*Cp*100 C=10.5g*1mol/18.016g*6.009 kJmol-1 Cp(metal) = 0.70 JK-1g—1.
9.
If the enthalpy change for the combustion of propane is -2220 kJ/mol propane, what quantity is
released when 1 kg of propane is burned?
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) = -2220 kJ
2220kJ * 1000/44 = 50,450 kJ
10.
Using the following thermochemical data, calculate the molar heat of combustion H of
methane, CH4:
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
H = -891kJ
2 CH4(g) + 3 O2(g)  2 CO(g) + 4 H2O(l)
H = -1215 kJ
2 C(s) + O2(g)  2 CO(g)
H = -221 kJ
C(s) + O2(g)  CO2(g)
H = - 394 kJ
11.
Calculate the standard molar enthalpy of formation of methane from the data given in question
10, your answer to question 10 and the following: Hf (H2O(l)) = -286 kJ/mol
H reaction = ( Hf (products) - Hf (reactants)
H = -891kJ Hf (H2O(l)) = 286 kJ/mol Hf (CO2(g)) = -394 kJ/mol
-891 = [(-394+(2*-286)) – ( Hf (methane)] Hf (methane) = +891-394+(2*-286) = -75kJ.mol
12.
When ammonia is oxidized to nitrogen dioxide and water, the quantity of heat released equals
349 kJ per mol of ammonia: 2 NH3(g) + 7/2 O2(g)  2 NO2(g) + 3 H2O(g)
H = -698 kJ
Calculate the standard molar enthalpy of formation of ammonia if:
O2 Hf =0
Hf (H2O(l)) = -286 kJ/mol and Hf (NO2(g)) = +33 kJ/mol
Hf (ammonia)=-47kJ/mol
13.
A 0.915 gram sample of sugar (C12H22O11; molar mass 342 g/mol) was ignited in a bomb
calorimeter in the presence of excess oxygen. Combustion was complete. The temperature of
the calorimeter and its contents rose by 3.53 C. If the heat capacity of the calorimeter and its
contents is 4250 JK-1, calculate the heat released per mole of sugar.
4250 JK-1 * 3.53 C ( same for C as K)= 15,000J
15,000J *342gmol-1/0.915g * 1 kJ/1000J = 5600 kJ/mol
14.
When 40 g of ammonium nitrate is dissolved in 100 g of water in a constant-pressure coffee
cup calorimeter, the temperature of the solution drops by 22.4 C. If the specific heat capacity
of the solution is 4.18 JK-1g-1, calculate the enthalpy of solution of ammonium nitrate.
mass of solution = 100+40=140 g.
4.18 JK-1g—1 * 140 g*22.5 C = 13,108K ( for 40 grams)
13,108 * 80.04/40 = 26,230 J = 26 kJ/mol
H solution = +26kJ/mol and endothermic process