CHEMISTRY - ENERGETICS
1. Determine the value of H for the following reaction:
Given: 2C(s) + H2(g) + 3Cl2(g) --> 2CHCl3(1)
2C(s) + 4H2(g) --> 2CH4(g)
1/2H2(g) + 1/2Cl2(g) --> HCl(g)
2.
CH4(g) + 3Cl2(g) --> CHCl3(1) + 3HCl(g)
H = -268.94 kJ
H = -149.62 kJ
 H = -92.31 kJ
HO(g) + Cl2(g) --> HOCl(g) + Cl(g) Find H for this reaction using Hess's law.
Given:
Cl2(g) --> 2Cl(g)
H = +242 kJ
H2O2(g) --> 2HO(g)
H = +134 kJ
2Cl(g) + H2O2(g) --> 2HOCl(g)
H = -210 kJ
3. Calculate the heat of formation Hfo of H2O2(1) from the following data:
H2(g) + 1/2O2(g) --> H2O(g)
 Ho = -241.818 kJ
2H(g) + O(g) --> H2O(g)
 Ho = -926.919 kJ
2H(g) + 2O(g) --> H2O2(g)
 Ho = -1070.60 kJ
2O(g) --> O2(g)
 Ho = -498.340 kJ
H2O2(1) --> H2O2(g)
 Ho = +51.46 kJ
Hint: Write a balanced equation for the heat of formation of one mole of H2O2(1) from its elements.
4. Calculate the heat of formation Hf of N2H4(1) from the following data:
2NH3(g) + 3 N2O(g) --> 4N2(g) + 3H2O(1)
 Ho = -1010 kJ
N2O(g) + 3H2(g)
--> N2H4(1) + H2O(1)
 Ho = -317.0 kJ
2NH3(g) + 1/2O2(g) --> N2H4(1) + H2O(1)
 Ho = -143.0 kJ
H2(g) + 1/2O2(g)
--> H2O(1)
 Ho = -286.0 kJ
Hint: Write the reaction for the formation of one mole at N2H4(1) from its elements
5. How much heat is required to increase the temperature of 380 g of diamond, which has a specific heat
of 0.502 J.g-1.K-1, from 30.0oC to 3000oC? Note: Diamond’s melting point is 3550oC.
6. When 6.22 kg of Cu absorbs 727.5 kJ , its temp goes from 20.5oC to 324.3oC. What’s the specific heat
7. How much heat would is needed to vaporize 98.7 g of CH3OH at its boiling point of 64.0oC Hvap 35.3 kJ/mol
8. 5.00 g samples of Fe and Au are both initially at 20.0oC, they are heated with 1.00 x 102 J of energy.
metal sample achieves the highest temperature? Fe = 0.444 J.g-1.K-1 and Au = 0.129 J.g-1.K-1.
Which
9. 550 grams of Al (c = 0.900 J.g -1 .K -1 at 80.0 o C is added to 675 mL of water (density = 1.00 g.mL -1 ) at a
temperature of 23.8oC. What is the final temperature?
10. Given:  HfoFe2O3(s) = -823.41 kJ.mol-1 and HfoFe3O4(s) = -1120.48 kJ.mol-1
(a) Calculate  Ho for the following reaction as written: 3Fe2O3(s) --> 2Fe3O4(s) + 1/2O2(g)
(b) Calculate  H per mole of Fe2O3 consumed and per mole of Fe3O4 formed.
(c) How many grams of Fe2O3(s) will decompose when 6.8781 kJ of heat are absorbed?
(d) How much heat in kJ must be absorbed when 2.3153 g of Fe3O4 is formed?
12.
0.6106 g of Benzoic acid C7H6O2 are burned in a bomb calorimeter containing 900.0 g of H2O. The heat
capacity of the bomb calorimeter itself is 303.4 J.K-1. The temperature rises from 24.02oC to 27.64oC. Find
the heat of combustion of benzoic acid in kJ.mol-1.
13.
The heat capacity of a bomb calorimeter can be determined by burning an accurate mass of a substance
whose heat of combustion (at constant volume) is known. When 0.7547 g of napthalene C10H8, was burned in a
bomb calorimeter containing 2000 g of water the temperature rose from 24.20oC to 27.62oC. What is the heat
capacity of the calorimeter if the heat of combustion of napthalene is -5150 kJ /mol?
14. A 100.0 mL sample of 0.400 mol.L-1 HCl(aq) is added to 150.0 mL of 0.300 mol.L-1 NH3(aq) in a calorimeter and
the temperature of the resulting solution rises from 23.92oC to 26.07oC.
(a) Determine the limiting reagent.
(b) Calculate the enthalpy change H for the following reaction in kJ.mol-1 for both reactants:
HCl(aq) + NH3(aq) --> NH4Cl(aq)
Additional data for problems 14 & 15, density of water = 1.00 g.mL-1
15.
The value of  H for the following reaction is determined in a constant pressure calorimeter:
CaO(s) + 2HCl(aq) --> CaCl2(aq) + H2O(1)
l.1216 g of CaO is added to 100 mL of 1.00mol.L-1 HCl, the temp rises from 24.94oC to 34.18 oC.
(a) Determine the limiting reagent.
(b) Find the value of  H in kJ.mol-1 for both reactants.
(c) What assumptions must be made?
16. Given the following information solve for Hfo H2S(g)
 HfoHNO3 = -174.10 kJ.mol-1  H oNO(g) = +90.25 kJ.mol-1  H
f
f water(g) = - 242KJ/mol
3H2S(g) + 2HNO3(1) ----> 2NO(g) + 4H2O(1) + 3S(s)
Ho = -552.7kJ
17. An ice calorimeter measures heats of reaction at constant volume. A sealed bomb containing 5.40 g Al and 19.89
g CuO is placed in an ice calorimeter containing 6.427 kg of ice and 6.692 kg of H2O(1) at 0oC.
After the
reacton it is found that 6.126 kg of ice and 6.993 kg of H 2 O(1) remain after melting stops.
2Al(s) + 3CuO(s) --> Al2O3(s) + 3Cu(s)
(a) Determine the limiting reagent
(b) How much ice melts?
(c) How much heat in kJ is needed to melt the ice in part (b)
(d) If all the heat of reaction (qr) is used to melt ice, calculate H per mole of the limiting reagent.
(e) Find H for the reaction as written.
18. The boiling point and freezing point of ethyl alcohol, C2H5OH, are 78.7oC and -117.3oC respectively. The
specific heat of ethyl alcohol(1) = 2.46 J.g .K-1. The heat of vaporization  H and the heat of fusion H
are 39.3
FUS
kJ.mol-1 and 7.61 kJ.mol respectively. How many much heat is needed to change 75.0 g of solidC2H5OH at -117.3
oC to gaseous C H OH at 78.7oC?
2 5
19. Write balanced thermochemical equations for the following. Include H as a reactant or product.
(a) heat of formation of K2CO3(s)
Hfo = -1146kJ.mol-1
(b) heat of neutralization of H2S(aq)
H = -32.4 kJ.mol-1 (using NaOH(aq) as the base)
(c) heat of solution of NH4Cl(s)
H = +15.13 kJ.mol-1
(d) heat of combustion of C2H6(g)
H = -1371 kJ.mol-1
-1
(e) heat of fusion of BaBr2
H = +25.1kJ.mol
2
Answers #1 H = -336.6 kJ #2. H = +70kJ #3. H= -187.79 kJ #4.H= +50.5 kJ #5. 5.67x10 kJ #6 .385
J.g-1.K-1 #7. 109 kJ #8 T (Fe) = 65.0oC T (Au) = 175oC #9 32.0oC 10 (a) Ho = +229.2 kJ
F
F
(b) 76.4 kJ.mol-1 Fe2O3 114.6 kJ.mol-1 Fe3O4 (c) 14.38 g (d) + 1.146 kJ #11 (a) -9098 kJ
(b) -4549 kJ.mol-1 B5H9 -1820 kJ.mol-1 B2O3 (c) 2.18 g
#12 -2.945 x 103 kJ.mol-1 #13 499 J.K-1
#14 56.2 kJ.mol-1 HCl -56.2 kJ.mol-1 NH3 #15 -194 kJ.mol-1 CaO -97.0 kJ.mol-1 HCl #16 -20.6 kJ.mol-1 #17 (b)
301 g ice (c) 100 kJ (d) -401 kJ.mol-1 CuO
(e) -1.20 x 103 kJ #18 1.12 x 105 J