Unit 4
Thermochemistry Worksheet
1.
The enthalpy of formation for silver nitrate is –2304 kJ/mol.
a. Write the balanced reaction representing this enthalpy.
b. What is the value of the enthalpy to form three (3) moles of silver nitrate?
c. What is the value of the enthalpy to form 15.490 grams of silver nitrate?
Answer:
a. Ag(s) + ½ N2(g) + 3/2 O2(g) → 1 AgNO3(s); ∆H = –2304 kJ/mol
b. (3 mol)(–2304 kJ/mol) = –6912 kJ
15.490 g  1mol   2304kJ 
c.


  210.08kJ
 169.88 g   1mol 
2.
When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na2SO4
solution at 25.0°C in a calorimeter, the white solid BaSO4 forms and the temperature of the
mixture increases to 28.1°C. Assuming that the calorimeter absorbs only a negligible quantity of
heat, that the specific heat capacity of the solution is 4.18J/°C•g, and that the density of the final
mixture is 1.0 g/mL, calculate the enthalpy change per mole of NaNO3 formed.
Answer:
1 Ba(NO3)2 + 1 Na2SO4 → 2 NaNO3 + 1 BaSO4
qmixture = mmixturecmixture∆tmixture = (1000+1000 g)(4.18)(28.1–25.0˚C) = +25916 J
Select one of the reactants to obtain moles:
(1.00 L)(1.00 M) = 1.00 mol Ba(NO3)2

  1mol Ba  NO3 2 
25916J

 
  12958 J
 1mol Ba  NO3 2   2mol Ba  NO3 2 
1.30x104 J
3.
The molar heat capacity of mercury is 28.1 J/mol·K. What is the specific heat capacity of this
metal?
Answer:
Specific heat capacity has units of J/g·˚C.
Since the temperature is a change in temperature, rather than a direct temperature, the unit
is based on the difference between two temperatures. Since ˚C and K are converted
through addition or subtraction, the difference between two temperatures in ˚C is the
same difference between the same temperatures in K. Take for instance Ti = 25˚C and
Tf = 55˚C. The difference between these two is Tf – Ti = 55 – 25 = 30˚C. At the same
time, these temperatures in Kelvin would appear as Ti = 298K and Tf = 328K. The
difference between these two is Tf – Ti = 328 – 298 = 30K. The magnitude is the same.
Therefore, the molar heat capacity of mercury can also be represented as J/mol·˚C.
Now, the moles can be converted to grams:
 28.1J   1mol Hg 
J
  0.140 g  C




mol
C
200.59
g
Hg



4.
Calculate the amount of heat absorbed by a 3.46 g sample of aluminum when the temperature
raises from 25.0˚C to 45˚C. (Specific heat of aluminum = 0.902 J/g•˚C) Is this endothermic or
exothermic? Why?
Answer:
q = (3.46 g)(0.902 J/g•˚C)(45˚C – 25˚C)
q = +62 J
endothermic, heat was absorbed by the water
5.
How many joules are required to raise the temperature of 0.500 kg of liquid water by 24.0˚C?
How many calories are needed?
Answer:
Temperature was “raised” so the ∆T = +24˚C
q = (500. g)(4.184 J/g•˚C)(+24˚C)
q = +50200 J
50200J 1cal

4.184J
1.20x10 4 cal
6.
The mass of a sample of potassium bromide in a beaker is 24.89 g. (The beaker itself weighs
23.14 g.) The potassium bromide was then dissolved in 100.00 g of distilled water, raising the
temperature from 20.05˚C to 25.40˚C. Assume all of the heat was absorbed by the water.
a. Write a balanced equation of the dissolving potassium bromide.
b. Calculate q of the reaction.
c. Is the solution process exothermic or endothermic?
d. What is q when 1.00 mole of potassium bromide is dissolved?
Answer:
a.
KBr(s) → K+(aq) + Br–(aq)
b. qH2O = (100.00 g)(4.184 J/g•˚C)(25.40 – 20.05 ˚C)
qH2O = +2240 J = –qKBr-dissolving
–qKBr-dissolving = –2240 J
c.
exothermic – whatever was absorbed by the water was released by the solution
process.
2240J 119g

175g 1mol 152000J / mol
d.
7.
A 4.00 g sample of salicylic acid, C7H6O3, is burned in a bomb calorimeter with a heat capacity
of 2.046 x 104 J/˚C. The combustion produces 87.7 kJ of heat, and the temperature after
combustion is 26.10˚C. Calculate the initial temperature of the bomb.
Answer:
87700 J = (2.046x104 J/˚C)(26.1 – Ti)
Ti = 21.8˚C
8.
A copper penny with a mass of 3.14 g is heated to 100.0˚C in boiling water. It is quickly dried
and placed into a Styrofoam cup containing 10.0 g of water at 25.0˚C. Copper has a specific
heat of 0.387 J/g•˚C. What will be the final temperature of the penny and the water? (Assume
that a negligible amount of heat is absorbed by the cup.)
Answer:
qCu = – qH2O
(3.14 g)(0.387 J/g•˚C)(Tf – 100.0˚C) = – (10.0 g)(4.184 J/g•˚C)(Tf – 25.0˚C)
1.215Tf – 121.5 = –41.84Tf + 1046
43.055Tf = 1167.5
Tf = 27.1 ˚C
9.
When 2.14 g of ethylene, C2H4(g), burns in a bomb calorimeter, the temperature rises 12.94˚C.
In a separate experiment, it is found that the addition of 6.23 kJ of heat to the calorimeter raises
the temperature by 1.61˚C.
a. What is the heat capacity of the calorimeter?
b. What is q for the combustion of one gram of ethylene?
c. What is q for the combustion of two moles of ethylene?
Answer:
a.
6230 J = Ccalorimeter (1.61˚C)
Ccalorimeter = 3870 J/˚C
b. qethylene = – qcalorimeter
qethylene = – (3870 J/˚C)(12.94˚C)
qethylene = –50100 J for 2.14 grams of ethylene
50100J
qethylene = 2.14g = –23400 J/g
c.
23400J 28.06g 2mol

g
1mol
1310000J  1310kJ
10. On a hot day, you take a six-pack of soda on a picnic, cooling it with ice. The metal in each
aluminum can weighs 38.5 g and contains 12.0 oz of soda. The specific heats of aluminum and
soda are 0.902 J/g·˚C and 4.10 J/g·˚C, respectively.
a. How much heat must be absorbed from the six-pack to lower the temperature from 25.0˚C to
5.0˚C?
b. How many pounds of ice must be melted to absorb this amount of heat? (∆Hfusion of ice = 6.00
kJ/mol.)
Answer:
a.
–qcooling = qcans + qsoda
mass of aluminum = (38.5 g)(6 cans) = 231 g
12.0g 1lb 453.6g 6cans
 2041.2g
16oz 1lb
mass of soda =
–qcooling = (231 g)(0.902 J/g•˚C)((5.0 – 25.0˚C) + (2041.2 g)(4.10 J/g•˚C)(5.0 –
25.0˚C)
–qcooling = –4167.24 –167378.4
qcooling = +172000J = +172 kJ (absorbed)
b.
172kJ 1mol 18.02g 1lb

6.00kJ 1mol 453.6g 1.14 lb of ice
11. A metal specimen with a mass of 25.467 g was heated to 100.0˚C in boiling water. The sample
was quickly dried and placed in a Styrofoam cup that contained 15.0 g of H2O having a
temperature of 24.3˚C. The mixture was stirred quickly and the temperature of the water rose to
31.2˚C. Calculate the specific heat of the metal.
Answer:
qmetal = – qH2O
(25.467 g)(c)(31.2 – 100.0˚C) = –(15.0 g)(4.184 J/g•˚C)(31.2 – 24.3˚C)
–1752(c) = –433.044
c = 0.247 J/g•˚C
12. Consider the following equations:
(1) CH3OH(l) + O2(g) → HCHO2(l) + H2O(l)
(2) CO(g) + 2 H2(g) → CH3OH(l)
(3) HCHO2(l) → CO(g) + H2O(l)
∆H˚ = –411 kJ
∆H˚= –128 kJ
∆H˚ = –33 kJ
Suppose Equation (1) is reversed and divided by 2, Equation (2) and (3) are multiplied by ½ and
then the three adjusted equations are added. What is the net reaction and what is the value of
∆H˚ for the net reaction?
Answer:
1
/2 HCHO2(l) + 1/2 H2O(l) → 1/2 CH3OH(l) + 1/2 O2(g)
∆H˚ = (–1/2)(–411 kJ)
1
/2 CO(g) + 1 H2(g) → 1/2 CH3OH(l)
∆H˚= (1/2)(–128 kJ)
1
/2 HCHO2(l) → 1/2 CO(g) + 1/2 H™O(l)
∆H˚ = (1/2)(–33 kJ)
1
HCHO2(l) + H2(g) → CH3OH(l) + /2 O2(g)
∆Hrxn = +125 kJ
13. Given the following thermochemical equations,
2 Cu(s) + S(s) → Cu2S(s)
S(s) + O2(g) → SO2(g)
Cu2S(s) + 2 O2(g) → 2 CuO(s) + SO2(g)
∆H˚ = –79.5 kJ
∆H˚ = –297.0 kJ
∆H˚ = –527.5 kJ
Calculate the value of the standard enthalpy of formation (in kilojoules) for 25.00 g of CuO(s).
Answer:
Cu2S(s) + 2 O2(g) → 2 CuO(s) + SO2(g)
∆H˚ = –527.5 kJ
∆H˚ = –79.5 kJ
2 Cu(s) + S(s) → Cu2S(s)
∆H˚ = (–1)(–297.0 kJ)
SO2(g) → S(s) + O2(g)
2 Cu + 1 O2 → 2 CuO
∆Hrxn = –310 kJ
310kJ 1mol 25.0g

2mol 79.55g
–48.8 kJ for 25.00 grams
14. Important reactions in the production of ozone, O3, in polluted air are
(1) 2 NO(g) + O2(g) → 2 NO2(g)
(2) NO2(g) + O2(g) → O3(g) + NO(g)
(3) O2(g) + O(g) → O3(g)
Use the data of ∆Hf˚ in your textbook’s appendix to calculate ∆H˚ (in kilojoules) for reactions
(1), (2), and (3). (The ∆Hf˚ for O(g) is +249 kJ/mol)
Answer:
(1) ∆Hrxn = ∑∆Hf(products) – ∑∆Hf(reactants)
∆Hrxn = [2(+33.9)] – [2(+90.4) + 0] = –113 kJ
(2) ∆Hrxn = ∑∆Hf(products) – ∑∆Hf(reactants)
∆Hrxn = [90.4 + 143] – [33.9 + 0] = +199.5 kJ
(3) ∆Hrxn = ∑∆Hf(products) – ∑∆Hf(reactants)
∆Hrxn = [143)] – [0 + 249] = –106 kJ
15. Given the following thermochemical equations,
2 H2(g) + O2(g) → 2 H2O(l)
N2O5(g) + H2O(l) → 2 HNO3(l)
½ N2(g) + 3/2 O2(g) + ½ H2(g) → HNO3(l)
∆H˚ = –571.5 kJ
∆H˚ = –76.6 kJ
∆H˚ = –174 kJ
Calculate ∆H˚ for the reaction
2 N2(g) + 5 O2(g) → 2 N2O5(g)
Answer:
2 H2O(l) → 2 H2(g) + O2(g
4 HNO3(l)) → 2 N2O5(g) + 2 H2O(l
2 N2(g) + 6 O2(g) + 2 H2(g) → 4 HNO3(l)
2 N2(g) + 5 O2(g) → 2 N2O5(g)
∆H˚ = (–1)(–571.5 kJ)
∆H˚ = (–2)–(76.6 kJ)
∆H˚ = (4)(–174 kJ)
∆Hrxn = +28.7 kJ