How to calculate enthalpy of combustion with examples.

This article involves enthalpy change of reaction, enthalpy change of formation, enthalpy of combustion and calculations involving conbustion.
Question 1
(a) State Hess's law
(b) Given the following data;
     C(s) + O2(g) --> CO2(g)   ΔH= -393 KJ/mol
      H2(g) + ½O2(g) --> H2O(l)    ΔH= -286 KJ/mol
      3C(s) + 4H2(g) --> C3H8(g) ΔH= -286 KJ/mol
Calculate the standard enthalpy of combustion of propane.

Answer:
(a) Hess's law states that the total heat change accompanying a chemical reaction is independent of the path taken. this law follows the first law of thermodynamics. its application lies in the calculation of change in enthalpy (ΔH) values which are difficult to obtain experimentally.
Hess's law is a concequence of the fact that enthalpy is a state function.

(b) writing out the equation of combustion of propane;

             C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(l)

using the formula below to obtain the standard heat of combustion;


ΔHc(C3H8) = 3 X ΔHf(CO2) + 4 X ΔHf(H2O) - ΔHf(C3H8) – 5 X ΔHf(O2)
                     = 3 X (-393) + 4X (-286) - (-104)- 5X (0)
                     = -1179 - 1144 + 104
                     = -2323 + 104
                     = -2219
                     = -2220vKJ/mol
Therefore, the standard heat of combustion is = -2220KJ/mol



Question 2
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 3
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]
( ΔHc(C2H6) = 2 X ΔHf(CO2) + 3 X ΔHf(H2O) - ΔHf(C2H6) – 3½ X ΔHf(O2)
( 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)
( ΔHc(C2H6) = - 1559.7 kJ mol-1




Question 4
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]
( ΔHc(C2H2) = 2 X ΔHf(CO2) + ΔHf(H2O) - ΔHf(C2H2) – 2½ X ΔHf(O2)
( ΔHc(C2H2) = 2 X(-393.5 kJ mol-1) + (-285.8 kJ mol-1) – (227 kJ mol-1) - 2½ X (0 kJ mol-1)
( ΔHc(C2H2) = -1299.8 kJ mol-1



Question 5
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]
( ΔHc(C3H8) = 3 X ΔHf(CO2) + 4 X ΔHf(H2O) - ΔHf(C3H8) – 5 X ΔHf(O2)
( -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)
( -2220 kJ mol –1 = - 1180.5 kJ mol-1 - 1143.2 – ΔHf(C3H8) – 0
( ΔHf(C3H8) = - 1180.5 kJ mol-1 - 1143.2 kJ mol-1 + 2220 kJ mol –1
( ΔHf(C3H8) = - 103.7 kJ mol-1



Question 6
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 7
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 8
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]
( ΔHc(C2H2) = 2 X ΔHf(CO2) + ΔHf(H2O) - ΔHf(C2H2) – 2½ X ΔHf(O2)
( ΔHc(C2H2) = 2 X(-393.5 kJ mol-1) + (-285.8 kJ mol-1) – (227 kJ mol-1) - 2½ X (0 kJ mol-1)
( ΔHc(C2H2) = -1299.8 kJ mol-1



Question 9
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]
( ΔHc(CH3OH) = ΔHf(CO2) + 2 X ΔHf(H2O) - ΔHf(CH3OH) – 1½ X ΔHf(O2)
( -715 kJ mol –1 = -393.5 kJ mol-1 + 2 X (-286.5 kJ mol-1) - ΔHf(CH3OH) - 1½ X (0 kJ mol-1)
( -715 kJ mol –1 = -393.5 kJ mol-1 -571.6 kJ mol-1 – ΔHf(CH3OH) – 0
( ΔHf(CH3OH) = -393.5 kJ mol-1 + -571.6 kJ mol-1 + 715 kJ mol –1
( ΔHf(CH3OH) = -250.1 kJ mol-1