NOTE: Since the enthalpy of vapourisation of water is 40.7 kJmol^-1, 3 x 40.7 kJ would be released when the water condenses, giving an enthalpy of combustion of -1442 -(3 x 40.7) = -1564 kJmol^-1, which is very close to the standard value. See Questions 3, 4 and 8 for examples of comparing standard enthalpy of reaction values with those calculated from bond enthalpies.

ΔH_comb(corrected) = ΔH_comb(via bond enthalpies) + ΔH_vap(cyclohexane) - 6 x ΔH_vap(water)

ΔH_comb(corrected) = -3720 + 30.0 - 6 x 40.7 = -3934.2 kJmol^-1

(d) Calculation (a) gave a value of -3915.6 kJmol^-1 for the standard enthalpy of combustion of cyclohexane (298K/1atm).

(a) Apart from (a) none of the answers can be considered as a standard enthalpy values at 298K/1 atm, since bond enthalpy calculations can only involve gaseous species and the fact that bond energies are based on average/typical values for similar molecules.

(b) The value of -3720 kJmol^-1, ignoring state changes, has an error of 195.6 kJmol^-1 compared to the standard enthalpy value based on standard enthalpy changes calculated in (a). This is ~5% error, hardly insignificant!

(c) When corrections are applied to take into account state changes with respect to 298K/1 atm. the error is reduced to 18.6 kJmol^-1 (~0.5%), and not a bad result from typical/average bond enthalpies in organic molecules

ΔH_comb(via bond enthalpies) + ΔH_vap(ethanoic acid) - 2 x ΔH_vap(water)

ΔH_comb(corrected) = -932 +51.6 - 2 x 40.7 = -3934.2 kJmol^-1 = -961.8 kJmol^-1

(d) Unlike in Q3, the corrected bond energy enthalpy value does not give a more accurate one, than that based solely on the bond enthalpies of gaseous species. In fact, applying the correction, actually makes the value even further from the standard enthalpy change calculated in (a)

I'm not sure why these bond enthalpy calculations are so 'inaccurate' (~7% and ~11%), but there is some uncertainty in the bond energies eg I could not find C-O and C=O bond energies for a carboxylic acid and this easily make a significant difference.

 C₂H₆(g) + I₂(s) ==> C₂H₅I(l) + HI(g)

ΔH^θ_{reaction,298K} = {ΔH^θ_f,298(iodoethane) + ΔH^θ_f,298(hydrogen iodide)} - (ΔH^θ_f,298(ethane)

ΔH^θ_{reaction,298K} = -39.1 +25.9 +84.7 = +71.5 kJmol^-1

(b)

Endothermic changes

to atomise one mole of iodine = 2 x +107 = +214 kJ (remember atomisation refers to 1 mol atoms)

breaking one C-H bond = +412 kJ

total = +626 kJ

Exothermic changes

one C-I bond formed = -228

one HI bond formed = -299

condensation of 1 mole iodoethane = -32

total = -559

Overall enthalpy change ΔH_reaction = +626 -559 = +67 kJmol^-1

(c) The two values are reasonably close with an error of 4.5kJ in (b) compared to the standard enthalpy of reaction value.

NOTE: The error of ~6% is quite significant BUT if any of the bond energy values are out by a few kJ, then an error of that magnitude readily results.

Don't forget to change the positive signs for bond enthalpies of dissociation into negative values of bond formation.

ΔH_sub(C(s)) + 4 x ΔH_{bond diss(H2(g))}

= (3 x 715) + (4 x 436)

= 2145 +  1744 = +3889 kJ

2 x ΔH_{bond form(C-C(g))} + 8 x ΔH_{bond form(C-H(g))}

(2 x -348) + (8 x -412) =

= -696 - 3296 = -3992 kJ

C–H single bond is 412, O=O double bond (in oxygen) is 496,

C=O double bond is 803 (in carbon dioxide) and H–O single bond is 463

Calculate the enthalpy of combustion of the reaction

methane + oxygen ==> carbon dioxide + water

CH_4(g) + 2O_2(g) ==> CO_2(g) + 2H₂O_(g)

  using displayed formulae

bonds broken and heat energy absorbed from surroundings, endothermic change

(4 x C–H) + 2 x (1 x O=O) = (4 x 412) + 2 x (1 x 496) = 1648 + 992 = 2640 kJ taken in

bonds formed and heat energy released and given out to surroundings, exothermic change

(2 x C=O) + 2 x (2 x O–H) = (2 x 803) + 2 x (2 x 463) = 1606 + 1852 = 3458 given out

The overall energy change is:

• 3338 – 2640 = 818 kJ/mol given out per mole methane burned,

• since more energy is given out than taken in, the reaction is exothermic.

• Energy change = ΔH^θ_c (methane) = –818 kJ/mol

Q11 Given the standard enthalpies of formation at 298K and 1 atm/101 kPa pressure in kJmol^-1, use a Hess's Law cycle to calculate the enthalpy of combustion of propan-1-ol

ΔH^θ_f (propan-1-ol) = -302.7; ΔH^θ_f (carbon dioxide) = -393.5  and  ΔH^θ_f (water) = -285.8;

Construction of the Hess's Law Cycle to theoretically calculate the enthalpy of combustion of propan-1-ol
CH3CH2CH2OH(l) + 4.5O2(g)  == ΔHcf (propan-1-ol)  ==> 3CO2(g) + 4H2O(l)
-ΔHθf (propan-1-ol)				3 x ΔHθf (carbon dioxide) + 4 x ΔHθf (water)
3C(s)  +  4H2(g) + 5O2(g)

Watch out for the minus sign on the left arrow ΔH.

Therefore using the principle of a Hess's Law cycle

ΔH^θ_c (propan-1-ol)  =  -ΔH^θ_f (propan-1-ol)  +  3 x ΔH^θ_f (carbon dioxide)  + 4 x ΔH^θ_f (water)

ΔH^θ_c (propan-1-ol)  =  -302.7  +  3 x -393.5  +  4 x -285.8  = ? kJmol^-1

ΔH^θ_c (propan-1-ol)  =  -302.7  +  1180.5  +  1143.2  = 2021 kJmol^-1  (to 4 sig. figs.)

Q12 Using the average bond enthalpies listed below, (a) calculate the enthalpy change for the complete combustion of propanol.

(b) Explain why this value differs from the standard enthalpy of combustion computed in Q4.2 (and in data tables).

• C-H = 413 kJ mol⁻¹
• C-C = 348 kJ mol⁻¹
• C-O = 358 kJ mol⁻¹
• O-H = 463 kJ mol⁻¹
• O=O = 498 kJ mol⁻¹
• C=O (in CO₂) = 805 kJ mol⁻¹
• O–H (in H₂O) = 463 kJ mol⁻¹

For (a) CH₃CH₂CH₂OH +  4.5O₂ ===> 3CO +  4H₂O _{(all assumed to be in a gaseous state)}

+ 4.5 O=O  ==> 3 O=C=O  +  4 H-O-H

(i) Bonds broken = (7 x C-H)  + (1 x C-O)  +  (1 x O-H)  +  (3.5 x O=O)

= (2 x C-C)  +  (7 x C-H)  +  (1 x C-O)  +  (1 x O-H)  +  (3.5 x O=O)

= (2 x 348)  +  (7 x 413)  + (358)  +  (463)  +  (4.5 x 498)

= (696)  +  (2891)  + (358)  +  (463)  +  (1743)

endothermic change = +6649 kJ

(ii) Bonds made = (6 x C=O)  +  (8 x O-H)

= (6 x C=O)  +  (8 x O-H)

= (4830)  +  (3704)

exothermic change = -8534 kJ

Since (ii) > (i) the reaction is overall exothermic

So ΔH^θ_c (propan-1-ol) = -1885 kJ mol^-1

(b) Reasons for difference in enthalpy of combustion values

(i) The calculation in (a) is based on all species being in a gaseous state, whereas for standard enthalpy of combustion values, the alcohol and water would be in the liquid state.

Energy would be absorbed to evaporate the alcohol, but energy would be released on the condensation of water.

(ii) The bond energies from data tables are based on average values, but accurate bond enthalpies for a specific bond vary slightly for different molecular context e.g. the C-O bond in propan-1-ol might be slightly different than in isomeric propan-2-ol or methoxyethane.

• Determining the energy change for a typical fuel combustion reaction

• 100 cm³ of water (100g) was measured into a simple calorimeter.

• The spirit burner contained the fuel ethanol C₂H₅OH ('alcohol') and weighed 18.62g at the start.

• The initial temperature of the water is taken.

• After burning some time, the flame is extinguished, the water stirred gently and the final water temperature is taken to get the temperature rise.

• The burner and fuel are then reweighed to see how much fuel had been burned.

• After burning it weighed 17.14g and the temperature of the water rose from 18 to 89^oC.

• The specific heat capacity of water is 4.2 Jg^-1K^-1

• Calculate the thermal energy change of combustion in J/g ethanol

• The temperature rise = 89 – 18 = 71^oC (exothermic, heat energy given out).

• Mass of fuel burned = 18.62–17.14 = 1.48g.

• Heat absorbed by the water = mass of water x SHC_water x temperature rise

  • = 100 x 4.2 x 71 = 29820 J (for 1.48g)

  • heat energy released per g = energy supplied in J / mass of fuel burned in g

  • heat energy released on burning = 29820 / 1.48 = 20149 J/g of C₂H₅OH

• Calculate the enthalpy change of combustion in kJmol^-1 of ethanol

  • Relative atomic masses A_r: C = 12, H = 1, O = 16

  • M_r(C₂H₅OH) = (2 x 12) + (1 x 5) + 16 + 16 = 46, so 1 mole = 46g

  • Heat released (given out) by 1 mole of C₂H₅OH = 46 x 20149 = 926854 J/mole or 927 kJ/mol (3 sf)

  • Enthalpy of combustion of ethanol = ΔH_combustion (ethanol) = –927 kJmol^–1

  • This means 927 kJ of heat energy is released on burning 46g of ethanol ('alcohol').

  • The data book value for the heat of combustion of ethanol is –1367 kJmol^–1, showing lots of heat loss in the experiment!

  • It is possible to get more accurate values by calibrating the calorimeter with a substance whose energy release on combustion is known.