Worked Solutions

Worked Solutions

Chapter 13

Question 51

An experiment was carried out to measure the heat of reaction of hydrochloric acid and sodium hydroxide, according to the equation HCl(aq) + NaOH(aq) → Na(aq) + H2O(l). Using a polystyrene cup as a calorimeter, 200 cm3 of each of the reactant 1 M solutions were mixed and the rise in temperature was found to be 13.7 kelvins.

(b) Calculate the heat of reaction.

Answer:

Heat change = mc ΔT

= 0.4 X 4.2 X 13.7

= 23.016 kJ

The number of moles of HCl in 200 cm3 of 1 M hydrochloric acid solution

= volume in litres X molarity

= 0.2 X 1

= 0.2 moles

Heat of reaction = - (23.016 / 0.2) = - 115.08 kJ mol-1

Question 64

Calculate the heat of combustion of ethane, as described in the equation C2H6(g) + 3½O2(g) → 2CO2(g) + 3H2O(l), given the heats of formation of ethane gas, carbon dioxide gas and water liquid are –84.7 kJ mol-1, -393.5 kJ mol-1 and –285.8 kJ mol-1 respectively.

Answer:

C2H6(g) + 3½O2(g) → 2CO2(g) + 3H2O(l) ΔHc = ?

ΔHr = Σ ΔHf[products] - Σ ΔHf[reactants]

Þ ΔHc(C2H6) = 2 X ΔHf(CO2) + 3 X ΔHf(H2O) - ΔHf(C2H6) – 3½ X ΔHf(O2)

Þ ΔHc(C2H6) = 2 X (-393.5 kJ mol-1) + 3 X (-285.8 kJ mol-1) – (- 84.7 kJ mol-1) – 2 X (0 kJ mol-1)

Þ ΔHc(C2H6) = - 1559.7 kJ mol-1

Question 65

Calculate the heat of combustion of ethyne, as described in the equation C2H2(g) + 2½O2(g) → 2CO2(g) + H2O(l), given the heats of formation of ethyne gas, carbon dioxide gas and water liquid are +227 kJ mol-1, -393.5 kJ mol-1 and –285.8 kJ mol-1 respectively.

Answer:

C2H2(g) + 2½O2(g) → 2CO2(g) + H2O(l) ΔHc = ?

ΔHr = Σ ΔHf[products] - Σ ΔHf[reactants]

Þ ΔHc(C2H2) = 2 X ΔHf(CO2) + ΔHf(H2O) - ΔHf(C2H2) – 2½ X ΔHf(O2)

Þ ΔHc(C2H2) = 2 X(-393.5 kJ mol-1) + (-285.8 kJ mol-1) – (227 kJ mol-1) - 2½ X (0 kJ mol-1)

Þ ΔHc(C2H2) = -1299.8 kJ mol-1

Question 66

The heat of combustion of propane, C3H8, as described in the equation C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l), is –2220 kJ mol-1 and the heats of formation of carbon dioxide gas and water liquid are –393.5 kJ mol-1 and -285.8 kJ mol-1 respectively. Calculate the heat of formation of propane.

Answer:

The equation for the heat of combustion of propane is

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔHc = -2220 kJ mol –1.

ΔHr = Σ ΔHf[products] - Σ ΔHf[reactants]

Þ ΔHc(C3H8) = 3 X ΔHf(CO2) + 4 X ΔHf(H2O) - ΔHf(C3H8) – 5 X ΔHf(O2)

Þ -2220 kJ mol –1 = 3 X (-393.5 kJ mol-1) + 4 X (-285.8 kJ mol-1) – ΔHf(C3H8) – 5 X (0 kJ mol-1)

Þ -2220 kJ mol –1 = - 1180.5 kJ mol-1 - 1143.2 – ΔHf(C3H8) – 0

Þ ΔHf(C3H8) = - 1180.5 kJ mol-1 - 1143.2 kJ mol-1 + 2220 kJ mol –1

Þ ΔHf(C3H8) = - 103.7 kJ mol-1

Question 98

When 100 cm3 of 0.5 M sulfuric acid solution, H2SO4, react with 100 cm3 of 1 M sodium hydroxide solution, NaOH, the temperature rises by 6.85 kelvins. Calculate the heat of reaction described by the equation

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

Answer:

Heat change = mc ΔT

= 0.2 X 4.2 X 6.85

= 5.754 kJ

The number of moles of H2SO4 in 100 cm3 of 0.5 M sulfuric acid solution

= volume in litres X molarity

= 0.1 X 0.5

= 0.05 moles

Heat of reaction = - (5.754 / 0.05) = - 115.08 kJ mol-1

Question 100

When 2.9 grams of butane gas, C4H10, are burned in excess oxygen, 143.85 kilojoules of heat are produced. Calculate the heat of combustion of butane.

Answer:

2.9 grams of butane = 2.9 / 58 moles = 0.05 moles

0.05 moles produces 143.85 kJ.

Therefore, 1 mole of butane produces 143.85 / 0.05 kJ = 2480.17 kJ

Heat of combustion of butane = -2877 kJ mol-1

Question 101

Calculate the heat of combustion of ethyne, as described in the equation C2H2(g) + 2½O2(g) → 2CO2(g) + H2O(l), given the heats of formation of ethyne gas, carbon dioxide gas and water liquid are +227 kJ mol-1, -393.5 kJ mol-1 and –285.8 kJ mol-1 respectively.

Answer:

C2H2(g) + 2½O2(g) → 2CO2(g) + H2O(l) ΔHc = ?

ΔHr = Σ ΔHf[products] - Σ ΔHf[reactants]

Þ ΔHc(C2H2) = 2 X ΔHf(CO2) + ΔHf(H2O) - ΔHf(C2H2) – 2½ X ΔHf(O2)

Þ ΔHc(C2H2) = 2 X(-393.5 kJ mol-1) + (-285.8 kJ mol-1) – (227 kJ mol-1) - 2½ X (0 kJ mol-1)

Þ ΔHc(C2H2) = -1299.8 kJ mol-1

Question 102

The heat of combustion of methanol, CH3OH, as described in the equation CH3OH (l) + 1½O2(g) → CO2(g) + 2H2O(l), is –715 kJ mol-1 and the heats of formation of carbon dioxide gas and water liquid are –393.5 kJ mol-1 and -285.8 kJ mol-1 respectively. Calculate the heat of formation of methanol.

Answer:

The equation for the heat of combustion of methanol is

CH3OH (l) + 1½O2(g) → CO2(g) + 2H2O(l) ΔHc = -715 kJ mol –1.

ΔHr = Σ ΔHf[products] - Σ ΔHf[reactants]

Þ ΔHc(CH3OH) = ΔHf(CO2) + 2 X ΔHf(H2O) - ΔHf(CH3OH) – 1½ X ΔHf(O2)

Þ -715 kJ mol –1 = -393.5 kJ mol-1 + 2 X (-286.5 kJ mol-1) - ΔHf(CH3OH) - 1½ X (0 kJ mol-1)

Þ -715 kJ mol –1 = -393.5 kJ mol-1 -571.6 kJ mol-1 – ΔHf(CH3OH) – 0

Þ ΔHf(CH3OH) = -393.5 kJ mol-1 + -571.6 kJ mol-1 + 715 kJ mol –1

Þ ΔHf(CH3OH) = -250.1 kJ mol-1

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