Question
Thymol has the formula C10H14O and its combustion can be written as: C10H14O(l) + 13 O2(g)...
Thymol has the formula C10H14O and its combustion can be written as:
C10H14O(l) + 13 O2(g) → 10 CO2(g) + 7 H2O(l)
The change in enthalpy of combustion of 1 mole of thymol at 298.15 K is determined to be -5660 kJ at constant pressure.
What is ΔE for this reaction at 298K?
Enter your answer to 4 significant figures in units of kJ. Be sure to include the sign.
Answers
The relationship between ∆H and ∆E is given by:
Which, if pressure is constant and gases can be considered as ideal can be transformed to:
Where ∆ngas is the change in gaseous moles, which is -3 for the reaction given (10 gaseous moles in proucts and 13 in reagents).
So, we can calculate:
ΔΕ = ΔΗ – pΔVΔΕ = ΔΗ - RTLηραςᎬᎫ AE = -5660kᎫ - 0.008314- , * 2Ꮽ8K * (-3moles) = -5653ᎬᎫ
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