Thermochem Practice

Question 1

Combustable materials are considered to be safer when they are more thermally stable. Fuels are harder to ignite, and are safer during accidents and long-term storage.

Fuel	H°f
Ethanol	–277.6 kJ/mol
Methane	–74.6 kJ/mol
Octane	–250.1 kJ/mol
Propane	–103.8 kJ/mol

Which of the following substances can be considered to be the most energetically stable?

A) methane
B) octane
C) ethanol
D) propane

Answer 1

Explanation:

The lower the heat of formation of a substance, the higher its energetic stability. In the example given, ethanol has a heat of formation of -277.6 kJ/mol - the lowest value of any substance in the table.

Question 2

An example of a change in only kinetic energy is

A) water changing from 25°C to 100°C.
B) uranium nuclei undergoing fusion.
C) methane burning.
D) ice melting at 0°C.

Answer 2

Explanation:

When a substance is heated or cooled, the kinetic energy of the particles increases or decreases with temperature. It stands to reason, therefore, that during a phase change, where the temperature of the substance remains constant as it changes state, the kinetic energy of the material remains constant. The energy added to the system to cause melting or boiling, or lost from the system to cause freezing or condensation, causes a change in potential energy as the particles of the substance rearrange themselves.

Question 3

Given:
Reaction 1:

CaCO3(s) CaO(s) + CO2(g)
ΔH = +180 kJ

Reaction 2:

CaO(s) Ca(s) + O2(g)
ΔH = +635 kJ

What is the heat of reaction for
2Ca(s) + O2(g) + 2CO2(g) 2CaCO3(s)

A) +815 kJ
B) -815 kJ
C) -455 kJ
D) -1630 kJ

Answer 3

Reaction 1 (× -1):
CaO(s) + CO2(g) CaCO3(s)
ΔH = -180 kJ

Reaction 2 (× -1):
Ca(s) + O2(g) CaO(s)
ΔH = -635 kJ

Add:
Ca(s) + O2(g) + CO2(g) CaCO3(s)
ΔHr = -815 kJ

Therefore:
2Ca(s) + O2(g) + 2CO2(g) 2CaCO3(s)
ΔHr = -1630 kJ

Question 4

Calcium carbide (CaC2) can be made by heating calcium oxide (lime) with carbon (charcoal).
CaO(s) + 3C(s) CaC2(s) + CO2(g)
ΔH = +464.8 kJ

How much heat is absorbed in a reaction in which 2.33 mol C(s) is consumed under these conditions?

A) 361 kJ
B) 36.0 kJ
C) 1.30× 103 kJ
D) 155 kJ

Answer 4

Heat absorbed:
= 2.33 mol C(s) × 464.8 kJ
3 mol C(s)
= 361 kJ

Question 5

When 50.0 mL of 0.200 mol/L NaOH(aq) reacts with 50.0 mL of 0.200 mol/L H2SO4(aq) in an insulated cup, the temperature rises by 2.00°C. Assume that no heat is lost to the surroundings and that for these aqueous solutions, the specific heat capacity is 4.184 J/g °C and the density is 1.00 g/mL.
What is the enthalpy of neutralization, ΔH, expressed in kJ/mol NaOH(aq)?

A) –0.867 kJ/mol
B) –42.8 kJ/mol
C) –156 kJ/mol
D) –83.7 kJ/mol

Answer 5

2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l)

mole of NaOH(aq):
= c × V
= 0.200 mol/L × 0.0500 L
= 0.0100 mol

mole H2SO4(aq):
= c × V
= 0.200 mol/L × 0.0500 L
= 0.0100 mol

0.0100 mol NaOH(aq) will react with:
0.0100 mol NaOH × 1 mol H2SO4
2 mol NaOH
= 0.00500 mol H2SO4(aq)

The NaOH(aq) is limiting and will be completely used up. 
total volume of mixture = 100.00 mL 
total mass of mixture = 100.00 g 
ΔT = 2.00°C

quantity of heat released: 
= Q 
= m × c × ΔT 
= 100.00 g × 4.184 J/g °C 2.00°C 
= 836.8 J

ΔH:
= – 0.8368 kJ
0.0100 mol
= -83.7 kJ/mol

Question 6

Given the reaction:
Cl2O(g) + 3OF2(g) 2ClF3(l) + 2O2(g) + 394 kJ
What is the heat of reaction, ΔHr, for the reaction
ClF3(l) + O2(g) Cl2O(g) + OF2(g)

A) +394 kJ
B) -131 kJ
C) -1182 kJ
D) +131 kJ

Answer 6

Explanation:

The given equation is exothermic and for this reaction ΔHr = -394 kJ.
The reaction ClF3(l) + O2(g) Cl2O(g) + OF2(g) is the reverse of the given equation divided by 3.

Heat of reaction for this reaction:
= × + 394 kJ
= +131 kJ

Question 7

The standard enthalpy of formation for H2O2(l) is –187.8 kJ/mol.
The equation that conveys this information is __________.

A) H2O(l) + O2(g) H2O2(l) + 187.8 kJ
B) H2O2(l) H2(g) + O2(g) + 187.8 kJ
C) H2(g) + O2(g) H2O2(l) + 187.8 kJ
D) H2O(l) + O2(g) H2O2(l) + 187.8 kJ

Answer 7

The heat of formation reaction will show H2O2(l) formed from its elements, H2(g) + O2(g), in an exothermic reaction.

Question 8

CH3OH(l) + O2(g) HCOOH(l) + H2O(g)
ΔfH(H2O(g)) = -241.8 kJ/mol
ΔfH(HCOOH(l)) = -425 kJ/mol
ΔfH(CH3OH(l)) = -239.2 kJ/mol

The enthalpy change for the reaction is

A) ΔHrxn = –471.6 kJ
B) ΔHrxn = +471.6 kJ
C) ΔHrxn = –427.6 kJ
D) ΔHrxn = +427.6 kJ

Answer 8

ΔHrxn:
= [ΔfH(H2O(g)) + [ΔfH(HCOOH(g))] – [ΔfH (CH3OH(l))]
= [(-241.8 kJ) + (-425 kJ)] – [(-239.2 kJ)]
= -427.6 kJ

Question 9

When the combustion of fuels, such as methane or wood, is compared to normal cellular respiration:

A) both processes produce CO2(g).
B) combustion is endothermic, while respiration is exothermic.
C) combustion is exothermic, while respiration is endothermic.
D) respiration must occur in the presence of light while combustion does not.

Answer 9

Consider the reactions of combustion to cellular respiration:

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l)

Question 10

Given that the standard enthalpy of formation for perchloric acid is –40.58 kJ/mol, the energy released when 2.00 mol of HClO4(l) is formed from its elements (to the nearest tenth) is ______ kJ.

Answer 10

^H= nHf
= (2 mol)(-40.58 kJ/mol)
=-81.2 kJ

^= Delta
*= formation

Question 11

2Na2O2(s) + 2H2O(l) 4NaOH(s) + O2(g)
ΔH = –109 kJ

calculate ΔH for the reaction:

NaOH(s) + O2(g) Na2O2(s) + H2O(l)

A) –27.3 kJ
B) +54.5 kJ
C) +27.3 kJ
D) +109 kJ

Answer 11

Explanation:

The ΔH for the reaction will be the opposite of the example reaction because it has been reversed. On top of this, the number of moles of reactant and product have been changed, so we need to account for this in order to calculate the final answer.

In the example reaction, ΔH = -109 kJ/mol for the reaction that yields 4 moles of NaOH. The reverse reaction uses just one mole of NaOH, so as we are using a quarter of the amount of substance we will yield a quarter of the energy. To calculate the answer, we first reverse the sign, then divide the answer by 4. -109 kJ/mol becomes 109 kJ/mol, then dividing by 4 gives a final answer of +27.3 kJ.

Question 12

Compounds A react to form Coupound D, either directly or via intermediate reactions.
A – ΔH1 –> D or
A – ΔH2 –> B – ΔH3 –> C – ΔH4 –> D

What does this illustrate?

A) the Law of Conservation of Mass
B) Le Chatelier’s principle
C) Hess’s Law
D) Activation energy

Answer 12

Explanation:

This is an example of Hess’ Law, which states that a reaction can proceed via direct or intermediate routes as long as the total enthalphy change is the same for each pathway.

Combustable materials are considered to be safer when they are more thermally stable. Fuels are harder to ignite, and are safer during accidents and long-term storage.

Question 13

Combustable materials are considered to be safer when they are more thermally stable. Fuels are harder to ignite, and are safer during accidents and long-term storage.

Fuel	H°f
Ethanol	–277.6 kJ/mol
Methane	–74.6 kJ/mol
Octane	–250.1 kJ/mol
Propane	–103.8 kJ/mol

Which of the following substances can be considered to be the most energetically stable?

A) methane
B) octane
C) ethanol
D) propane

Answer 13

Explanation:

The lower the heat of formation of a substance, the higher its energetic stability. In the example given, ethanol has a heat of formation of -277.6 kJ/mol - the lowest value of any substance in the table.

Question 14

The dilution of pure nitric acid, HNO3(l), can be represented by the following equation:

HNO3(l) HNO3(aq) + heat
Given:

ΔH°f (HNO3(l)) = -173 kJ/mol
ΔH°f (HNO3(aq)) = -207 kJ/mol

What quantity of heat is given off when 1.0 × 102 g of HNO3(l) is diluted to form a 1.0 mol/L aqueous solution?

A) 54 kJ
B) 3.8 kJ
C) 21 kJ
D) 3.3 × 102 kJ

Answer 14

Explanation:

ΔHr:
= ΔH°f (HNO3(aq)) - ΔH°f (HNO3(l))
= -207 kJ/mol -(-173 kJ/mol)
= -34 kJ/mol

molar mass HNO3 = 63.02 g/mol

mol HNO3:
= 100g/ 63.02 g/mol
= 1.59 mol

heat given off:
= 1.59 mol × 34 kJ/mol
= 54 kJ

(Note that the concentration of the diluted solution does not affect the quantity of heat given off.)

Question 15

A 2.05 g sample of a metal alloy wire is heated to 98.88°C. It is then quickly dropped into 28.0 g of water in a calorimeter. The water temperature rises from 19.73°C to 21.23°C. Calculate the specific heat of the alloy.

A) 11.4 J/g·°C
B) 0.263 J/g·°C
C) 1.10 J/g·°C
D) 0.867 J/g·°C

Answer 15

Explanation:

heat lost by metal alloy = heat gained by water

m × c × ΔT = m × c × ΔT
2.05 g × c × (98.88 °C -21.23°C) = 28.0 g × 4.184 J/g°C × (21.23°C – 19.73°C)
c = 1.10 J/g °C

Question 16

Methane can be broken down into its base elements as in the following reaction:
CH4(g) C(s) + 2H2(g)
In this decomposition, the

A) reactants have more kinetic energy that do the products.
B) products have more kinetic energy that do the reactants.
C) reactants have more potential energy that do the products.
D) products have more potential energy that do the reactants.

Answer 16

Explanation:

The enthalpy change for all chemical reactions is a change in potential energy. Kinetic energy changes result in a change in temperature. This eliminates answers A & B.

The enthalpy change for the decomposition of methane is the same as the enthalpy change for the formation reaction, but with opposite sign. This value can be found in a chemistry data booklet. For the decomposition of methane, the standard enthalpy change is +74.6 kJ/mol. Since the enthalpy change is positive, the products have more energy than the reactants. Answer D is correct.

Question 17

The decomposition of sulfur trioxide is represented by the equation
2SO3(g) 2SO2(g) + O2(g)
ΔH° = + 198 kJ

Based upon this information, which one of the following thermochemical equations is correct?

A) SO3(g) SO2(g) + O2(g) + 198 kJ
B) SO2(g) + O2(g) + 198 kJ SO3(g)
C) SO2(g) + O2(g) SO3(g) + 99 kJ
D) SO3(g) SO2(g) + O2(g) + 99 kJ

Answer 17

Explanation:

Since 2SO3(g)  2SO2(g) + O2(g) 
ΔH° = + 198 kJ

then, 2SO3(g) + 198 kJ 2SO2(g) + O2(g)

or 2SO2(g) + O2(g) 2SO3(g) + 198 kJ

divide through by 2:
SO2(g) + O2(g) SO3(g) + 99 kJ

Question 18

Calcium carbide (CaC2) can be made by heating calcium oxide (lime) with carbon (charcoal).
CaO(s) + 3C(s) CaC2(s) + CO2(g)
ΔH = +464.8 kJ

How much heat is absorbed in a reaction in which 2.33 mol C(s) is consumed under these conditions?

A) 361 kJ
B) 36.0 kJ
C) 1.30× 103 kJ
D) 155 kJ

Answer 18

Explanation:

Heat absorbed:
= 2.33 mol C(s) × 464.8 kJ /
3 mol C(s)
= 361 kJ

Question 19

The thermochemical equation for the formation of hydrogen fluoride gas from its elements is given by
H2(g) + F2(g) 2HF(g) + 546 kJ

From this equation, it can be stated that _____________.

A) ΔH°f (HF(g)) = - 546 kJ/mol
B) the total bond energy in H2(g) + F2(g) is 546 kJ less than the bond energy in 2HF(g)
C) the total enthalpy in H2(g) + F2(g) is 546 kJ higher than the enthalpy in 2HF(g)
D) the formation of HF(g) from its elements at standard conditions is endothermic

Answer 19

C)

Weaker bonds are less stable, have a higher energy and are more easily broken.

Question 20

Substance	ΔHr (kJ/mol)
KClO3(s)	-397.7
KCl(s)	-436.5
O2(g)	0
2KClO3(s)  2KCl(s) + 3O2(g)

What quantity of heat is released when 1.00 g of KClO3(s) decomposes?

A) 0.317 kJ
B) 38.8. kJ
C) 0.633 kJ
D) 77.6 kJ

Answer 20

Explanation:

ΔHr:
= [2 mol KCl × (-436.5 kJ/mol)] – [2 mol KClO3(s) × (-397.7 kJ/mol)
= -77.6 kJ per 2 mol KClO3(s)
= -38.8 kJ/mol KClO3(s)

mol KClO3(s):
= 1.00 g
122.55 g/mol
= 0.00816 mol

heat released:
= 38.8 kJ/mol × 0.00816 mol
= 0.317 kJ

Question 21

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) + 2043.9 kJ
When propane is burned in a camp stove, the molar heat of reaction is

A) –103.8 kJ/mol
B) –2219.9 kJ/mol
C) +2219.9 kJ/mol
D) –2043.9 kJ/mol

Answer 21

Explanation:

Remember that in a chemical equation we are looking at things from the point of view of the surroundings. If we are asked about the heat of reaction, we need to reverse this and see things from the point of view of the system. This means that if an equation states that 2000 kJ of energy are produced by the burning of a substance then from a thermodynamic point of view, 2000 kJ of energy have left the system. In this case, 2043.9 kJ of energy are produced by the reaction, therefore the heat of reaction is the opposite of this, i.e. -2043.9 kJ.

Question 22

When phenol (C6H5OH(s)) burns according to 2C6H5OH(s) + 14O2(g) 12CO2(g) + 6H2O(g) the enthalpy of reaction is –6135.8 kJ. The molar enthalpy of formation for phenol (to the nearest tenth) is __________ kJ/mol.

Answer 22

Explanation:
ΔHrxn = [6ΔfH (H2O(g)) + 12ΔfH(CO2(g))] – [2ΔfH (C6H5OH(s)]

-6135.8 = [6(-241.8 kJ) + 12(-393.5 kJ)] – [2ΔfH (C6H5OH(s)]

ΔfH [C6H5OH(s)] = -18.5 kJ/mol

Question 23

BaO(s) + CO2(g) BaCO3(s)
ΔH = –274.8 kJ
ΔfH(BaO(s)) = -558.1 kJ/mol

ΔfH(CO2(g)) = -393.5 kJ/mol

Based upon the reaction, the molar heat of formation for barium carbonate is

A) –548.1 kJ/mol
B) –274.8 kJ/mol
C) –1226.4 kJ/mol
D) +666.7 kJ/mol

Answer 23

Explanation:

ΔrxnH = [ΔfH (BaCO3(s))] – [ΔfH(CO2(g)) + [ΔfH (BaO(s))]

-274.8 kJ = [ΔfH (BaCO3(s))] – [(-393.5 kJ) + (-558.1 kJ)]

ΔfH (BaCO3(s)) = -1226.4 kJ/mol

Question 24

The dilution of pure nitric acid, HNO3(l), can be represented by the following equation: 
HNO3(l)  HNO3(aq) + heat
Given: 
ΔH°f (HNO3(l)) = -173 kJ/mol
ΔH°f (HNO3(aq)) = -207 kJ/mol

What quantity of heat is given off when 1.0 × 102 g of HNO3(l) is diluted to form a 1.0 mol/L aqueous solution?

A) 54 kJ
B) 3.8 kJ
C) 21 kJ
D) 3.3 × 102 kJ

Answer 24

Explanation:

ΔHr:
= ΔH°f (HNO3(aq)) - ΔH°f (HNO3(l))
= -207 kJ/mol -(-173 kJ/mol)
= -34 kJ/mol

molar mass HNO3 = 63.02 g/mol

mol HNO3:
= 100g/ 63.02 g/mol
= 1.59 mol

heat given off:
= 1.59 mol × 34 kJ/mol
= 54 kJ

(Note that the concentration of the diluted solution does not affect the quantity of heat given off.)