1. Energetics
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2. Question 1 A B C D Which statement about this reaction is correct?
2Fe(s) + 3CO2(g)  Fe2O3(s) + 3CO(g) ΔHƟ=+26.6kJ A
13.3kJ of energy are absorbed for every mole of Fe reacted. B
26.6kJ of energy are absorbed for every mole of Fe reacted. C
26.6kJ of energy are released for every mole of Fe reacted. D
53.2kJ of energy are released for every mole of Fe reacted.

3. Wrong Answer!
For 2 moles of Fe, 26.6kJ of energy is absorbed to break the bonds. 1 mole of Fe will require 13.3kJ of energy.
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4. Question 2
Which of the following is observed when the change in enthalpy is positive for the dissoving of a salt in water in an insulated copper beaker? A
Heat is evolved to the surroundings and the beaker feels cold. B
Heat is evolved to the surroundings and the beaker feels warm. C
Heat is absorbed from the surroundings and the beaker feels warm. D
Heat is absorbed from the surroundings and the beaker feels cold.

5. Wrong Answer!
A positve enthalpy change means that heat flows from the surroundings into the system.
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6. All the following processes are exothermic except
Question 3
All the following processes are exothermic except A
2C2H5(g)  C4H10(g) B
F2(g)  2F-(g) C
Cl(g) + e-  Cl-(g) D
4Fe(s) + 3O2(g)  2Fe2O3 (s)

7. Bond breaking is always endothermic.
Wrong Answer!
Bond breaking is always endothermic.
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8. Question 4
What can be deduced about the relative stability of the reactants and products and the sign of ΔH, from the enthalpy level diagram below?
Relative stability Sign of ΔH
reactants A
products more stable B
products more stable ΔH C
reactants more stable
products D
reactants more stable

9. Wrong Answer!
The vertical axis represents potential energy, energy or enthalpy. The reactants are high in energy and hence unstable; the products are lower in energy and hence more stable. The difference is the released in the form of heat energy.
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10. Question 5
What is the specific heat capacity of an alcohol in Jg-1K-1 if 560.0J of heat are required to raise the temperature of a 64.0g sample of ethanol from 295.0K to 310.0K? A
0.583 B
0.194 C
8.75 D
0.292

11. Wrong Answer!
Heat energy = mass x specific heat capacity x change in temp
560 = 64.0 x c x (310 – 295 )
c = Jg-1K-1
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12. Question 6
When mol of nitric acid is reacted with mol of potassium hydroxide in water, the temperature of the system increases by 13.70C. Calculate the enthalpy of reaction in kJmol-1
HNO3(aq) + KOH(aq)  KNO3(aq) + H2O(l)
Assume that the heat capacity of the system was 209.2J 0C-1. A
+57.3 kJmol-1 B
+2.87 kJmol-1 C
-2.87 kJmol-1 D
-57.3 kJmol-1

13. Wrong Answer! Try Again Heat energy = 13.7 x 209.2 = 2866.04J
Enthalpy of reaction per mole of KCl = (1/0.05) x = Jmol-1
Sign is negative because energy is given out.
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14. Question 7
The bond energies for H2 , I2 and HI are 432, 149 and 295 kJmol-1 respectively. From these data, what is the enthalpy change(in kJ) for the reaction below?
H2(g) + I2(g)  2HI(g) A
+286 B +9 C -9 D
-286

15. Enthalpy change = [(432 + 149)] – [(295 x 2)] = -9 kJ
Wrong Answer!
Enthalpy change = [( )] – [(295 x 2)] = -9 kJ
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16. Question 8 A B C D Consider the following equation:
6CO2(g) + 6H2O(l)  C6H12O6(s) + 6H2O(g)
ΔH=2824kJmol-1
What is the enthalpy change associated with the production of 100.0g of C6H12O6? A
157 kJ B
282 kJ C
508 kJ D
1570 kJ

17. Wrong Answer!
Amount of C6H12O6= (100.0g/180gmol-1) = 0.55 mol-1
Enthalpy change = 0.55 x 2824 = kJ
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18. Question 9 A B C D Metal Specific heat capacity (Jg-1K-1)
The specific heat capacities of some metals are given below.
Metal Specific heat capacity (Jg-1K-1)
copper
magnesium
mercury
platinium
If 100kJ of heat is added to 10.0g samplesof each of the metals above, which are all at 250C, which metal will have the lowest temperature? A
copper B
magnesium C
mercury D
platinium

19. Wrong Answer!
Q = mcΔT, q/mc = ΔT, hence if c is increased, then ΔT will decrease. A high heat capacity means more energy is required to raise the temperature.
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20. Question 10
The bond energy for the H-F bond is equal to the enthalpy change for which process? A
HF(g)  ½F2(g) + ½H2(g) B
½F2(g) + ½H2(g)  HF(g) C
HF(g)  H(g) + F(g) D
H+(g) + F-(g)  HF(g)

21. Wrong Answer!
The H-F bond enthalpy is the amount of energy (in kJ) required to break 1 mole of HF covalent bond into gaseous hydrogen and fluorine atoms (under standard thermodynamic conditions)
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22. Question 11
When a sample of a pure hydrocarbon (melting point 850C) cools, the temperature is observed to remain constant as it solidifies. Which statement accounts for this observation? A
The heat released in the change of state equals the heat loss to the surroundings. B
The temperature of the system has fallen to room temperature. C
The solid which forms insulates the system, preventing heat loss. D
Heat is gained from the surroundings as the solid forms, maintaining a constant temperature.

23. Wrong Answer!
During freezing intermolecular forces (van der Waals’ forces) are formed and heat energy is released to the surroundings.
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24. Question 12 A B C D Consider the following reactions:
N2(g) + 3H2(g)  2NH3(g)
Bond enthalpies (in kJmol-1) involved in the reaction are:
NΞN a
H-H b
N-H c
Which expression could be used to calculate the enthalpy of reaction? A
a + 3b - 2c B
a - 3b + 6c C
6c – a + 3b D
a + 3b – 6c

25. Wrong Answer!
Enthalpy change = ∑(bond broken) - ∑(bond made)
= (NΞN + 3 x H-H) – (6 x N-H)
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26. Question 13
The enthalpy changes for two different hydrogenation reactions of C2H2 are:
C2H2 + H2  C2H4 ΔHƟ1
C2H2 + 2H2  C2H6 ΔHƟ2
Which expression represents the enthalpy change for the reaction below?
C2H4 + H2  C2H6 ΔHƟ = ? A
ΔHƟ1 + ΔHƟ2 B
ΔHƟ1 - ΔHƟ2 C
ΔHƟ2 – ΔHƟ1 D
- ΔHƟ1 - ΔHƟ2

27. Wrong Answer!
Using the energy cycle by Hess law, the enthalpy change is obtained by reverting ΔHƟ1 and adding to ΔHƟ2
C2H4 + 2H2  C2H6 ΔHƟ
ΔHƟ ΔHƟ2
C2H2 + 2H2
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28. Question 14 A B C D 2KHCO3(s) K2CO3(s) + CO2(g) + H2O(l)
+ 2HCl(aq) HCl(aq)
2KCl(aq) + 2CO2(g) + 2H2O(l)
This cycle may be used to determine ΔHƟ for the decomposition of potassium hydrogen carbonate.
Which expression can be used to calculate the ΔHƟ ?
ΔHƟ2
ΔHƟ1 A
ΔHƟ = ΔHƟ1 + ΔHƟ2 B
ΔHƟ = ΔHƟ1 - ΔHƟ2 C
ΔHƟ = ½ ΔHƟ1 - ΔHƟ2 D
ΔHƟ = ΔHƟ2 – ΔHƟ1

29. Wrong Answer!
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30. Congratulations!