HELP NOTES for the extra ΔH Q's for Advanced Level Chemistry

Enthalpy Calculation Revision Question Answers

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Thermodynamics Part 1: Thermochemistry - Calorimetry experiments - Enthalpies of reaction, formation, combustion and bond dissociation are explained with exemplar calculations  *   EMAIL

 Original set of ∆H enthalpy of combustion/formation calculation questions to go with these answers

 GCSE/IGCSE notes on Energy Changes - exothermic/endothermic reaction changes

See also fully worked out examples from calorimeter data and using Hess's Law cycles to solve problems

 I DO MY BEST TO CHECK MY CALCULATIONS, as you yourself should do,  BUT I AM HUMAN! AND IF YOU THINK THERE IS A 'TYPO' or CALCULATION ERROR PLEASE EMAIL ME ASAP TO SORT IT OUT!

Answers to Questions 1 and 2

Q1(c)(ii) Apart from the fact that bond enthalpy values are typical/average values in a variety of similar molecules, the bigger source of error is that the calculation in (c) does NOT take into account the enthalpy of vaporisation of water. Extra energy would be released from the condensation of water, giving a more exothermic value ie closer to -1560 kJmol-1.

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.

topOriginal set of Questions  *  Thermochemistry and Enthalpy Notes

 

topOriginal set of Questions

 

Answers to Questions 3 and 4

Q3(c) to apply a 'correction' to your answer to Q3(b) you need to allow for the endothermic vapourisation of 1 mole of cyclohexane and the exothermic condensation of 6 moles of water, therefore ...

ΔHcomb(corrected) = ΔHcomb(via bond enthalpies) + ΔHvap(cyclohexane) - 6 x ΔHvap(water)

ΔHcomb(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

topOriginal set of Questions  *  Thermochemistry and Enthalpy Notes

 

Q4(c) ΔHcomb(corrected) = ΔHcomb(via bond enthalpies) + ΔHvap(ethanoic acid) - 2 x ΔHvap(water)

ΔHcomb(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.

topOriginal set of Questions  *  Thermochemistry and Enthalpy Notes

 

Answers to Questions 5 to 7

(c) doc b

topOriginal set of Questions  *  Thermochemistry and Enthalpy Notes

 

Answers to Question 8

Q8(a) ΔHθreaction,298K = ∑ΔHθf,298(products) - ∑ΔHθf,298(reactants)

 C2H6(g) + I2(s) ==> C2H5I(l) + HI(g)

ΔHθreaction,298K = {ΔHθf,298(iodoethane) + ΔHθf,298(hydrogen iodide)} - (ΔHθf,298(ethane)

ΔHθreaction,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 ΔHreaction = +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.

topOriginal set of Questions  *  Thermochemistry and Enthalpy Notes

 

Answers to Question 9

Q9   Calculating the enthalpy of formation of propane from bond energy data and ∆Hsub(carbon)

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

3C(s) + 4H2(g) (c) doc b C3H8(g)

ΔHsub(C(s)) + 4 x ΔHbond diss(H2(g))

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

= 2145 +  1744 = +3889 kJ

(c) doc b (c) doc b

2 x ΔHbond form(C-C(g)) + 8 x ΔHbond form(C-H(g))

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

= -696 - 3296 = -3992 kJ

3C(g) + 8H(g)
∆Hform(C3H8) = +3889 + (-3992) =  -103 kJ mol-1 (data book value -104 kJ mol-1)

(c) doc b

Original set of Questions  *  Thermochemistry and Enthalpy Notes

I DO MY BEST TO CHECK MY CALCULATIONS, as you yourself should do,  BUT I AM HUMAN! AND IF YOU THINK THERE IS A 'TYPO' or CALCULATION ERROR PLEASE EMAIL ME ASAP TO SORT IT OUT!

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See also fully worked out examples from calorimeter data and using Hess's Law cycles to solve problems

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