2.4 EXT Structure and Bonding practise questions: Energy

Question 1

When Iron Corrodes in moist air, rust is formed according to the following equation:
4Fe(s) + 3O₂(g) + 6H₂O(l) → 4Fe(OH)₃(s)
△ᵣH = -3310kJ mol⁻¹

Calculate the amount of energy when 1 mole of rust is formed.

Answer 1

When 4 moles of Fe(OH)₃ is formed, -3310kJ/mol of energy is released. Therefore, when 1 mole of rust is produced, the energy released is 3310/4 = -827.5kJ/mol = -828kJ/mol

Question 2

When Iron Corrodes in moist air, rust is formed according to the following equation:
4Fe(s) + 3O₂(g) + 6H₂O(l) → 4Fe(OH)₃(s)
△ᵣH = -3310kJ mol⁻¹

Calculate the mass of Iron needed to form 1 mole of rust.

Answer 2

If the product of the balanced equation is reduced 3/4, then the reactants must be reduced by 3/4. Therfore, for 1 mole of rust, 1 Fe atom is needed. M(Fe) = 55.89mol⁻¹ nxM=m = 1x55.89 = 55.89g

Question 3

When Iron Corrodes in moist air, rust is formed according to the following equation:
4Fe(s) + 3O₂(g) + 6H₂O(l) → 4Fe(OH)₃(s)
△ᵣH = -3310kJ mol⁻¹

Calculate the mass of rust formed when 6.52g of Iron corrodes.

Answer 3

Number of moles in 6.25g of Iron:
(m/M = n) 6.52g ÷ 55.8gmol⁻¹ =0.117moles

Molar mass of Fe(OH)₃:
M(Fe) +3M(O) + 3M(H) = M(55.8) + 3M(16.0) + 3M(1.01) = 106.8gmol⁻¹

Mass of rust(g):
nxM= m
0.117 x 106.8 =12.5

Question 4

When Iron Corrodes in moist air, rust is formed according to the following equation:
4Fe(s) + 3O₂(g) + 6H₂O(l) → 4Fe(OH)₃(s)
△ᵣH = -3310kJ mol⁻¹

Explain why a rusting piece of Iron doesn’t feel hot.

Answer 4

Although rusting is an exothermic reaction, the Iron doesn’t feel hot because it reacts with Oxygen very slowly, meaning the energy is relased slowly over a period of time and spreads out into the surroundings, so there is no noticiable increase in temperature to human skin.

Question 5

When 1 mole of of hexane (C₆H₁₄) is burnt in Oxygen, it releases 4163 kJ of energy.

Write a balanced thermochemical equation for the complete combustion of hexane.

Answer 5

C₆H₁₄ (s) + 9.5O₂ (g) → 6CO₂ (g) + 7H₂O

△꜀H = -4163 kJ/mol⁻¹

Question 6

When 1 mole of of hexane (C₆H₁₄) is burnt in Oxygen, it releases 4163 kJ of energy.

Calculate the amount of energy per gram of carbon dioxide gas released by the hexane gas.

Answer 6

1 mole of C₆H₁₄ produces 6 CO₂ moles.

M(C) + 2M(O) = M(12.0) + 2M(16.0 = 44gmol⁻¹

44gmol⁻¹ x 6 moles = 264g (nxM=m)

energy produced ÷ grams of CO₂ released = 4163÷264 = 15.8kJmol⁻¹

Question 7

When 1 mole of of hexane (C₆H₁₄) is burnt in Oxygen, it releases 4163 kJ of energy.

How much energy is released when 0.25 mole of hexane is burnt?

Answer 7

(C₆H₁₄) ÷ 4 = 0.25(C₆H₁₄)
4163 ÷ 4 =1040.75kJ

If one mole released 4163kJ of energy, the 1/4 gives us 1040.75kJ

Question 8

Ethane gas, C₂H₆, is one of the components of CNG. The enthalpy of combustion, △꜀H, for ethane is -1557 kJ mol⁻¹

Write a balanced thermonchemical equation for the complete combustion of ethane.

Answer 8

C₂H₆ (g) + 3.5O₂ (g) → 2CO₂ (g) +3H₂O (l)
△꜀H = -1557kJmol⁻¹

Question 9

Ethane gas, C₂H₆, is one of the components of CNG. The enthalpy of combustion, △꜀H, for ethane is -1557 kJ mol⁻¹

Calculate the amount of heat released for each kilogram of ethane burnt.

Answer 9

m = nxM
m= 1x (2M(12.0) + 6(1.01)) m= 30.06g

1kg= 1000g

moles of ethane = 1000 ÷ 30.06 = 32.3 moles

32.3 moles x 1557 kJ mol⁻¹ = 50291.1kJ

Question 10

Octane is a mojor component in petrol, and burns according to the following equation:

C₈H₁₈ (l) +12½O₂ (g) → 8CO₂ (g) +9H₂O(l)
△꜀H = -5500kJmol⁻¹

Is the burning of Octane an endothermic or exothermic reaction? Give a reason for your answer.

Answer 10

Exothermic because the change in enthalpy is negative, and all combustion reactions are exothermic.

Question 11

Octane is a mojor component in petrol, and burns according to the following equation:

C₈H₁₈ (l) +12½O₂ (g) → 8CO₂ (g) +9H₂O(l)
△꜀H = -5500kJmol⁻¹

1.00 Litre of Octane contains 6.12 moles of the fuel. Calculate the energy released when 1 litre of fuel is burnt.

Answer 11

1 mole of C₈H₁₈ produces -5500kJ mol⁻¹.
Therefore if both sides are multiplied by 6.12, 6.12 moles of C₈H₁₈ produces -33660kJmol⁻¹

Question 12

Octane is a mojor component in petrol, and burns according to the following equation:

C₈H₁₈ (l) +12½O₂ (g) → 8CO₂ (g) +9H₂O(l)
△꜀H = -5500kJmol⁻¹

Using Hydrogen gas, (H₂) as a fuel as a fuel for cars, rather than Octane, is viewed as better for the environment.

H₂ (g) + ½O₂ (g) → H₂O (l)
△ᵣH = -286kJmol⁻¹

Calculate the mass of Hydrogen required to produce -33660kJmol⁻¹ of energy. State your answer in (3sf)

Answer 12

Energy released by 1 litre of Octane ÷ Energy released by 1 mole of Hydrogen gas

-33660 ÷ 286 = 117 moles required

2M(H)= 2M(1.01) = 2.02 = 236 (3sf)