Thymol has the formula C10H14O and its combustion can be written as: C10H14O(l) + 13 O2(g)...

Question

Question

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.

science chemistry

Answer 1

Answer #1

The relationship between ∆H and ∆E is given by:

ΔΕ = ΔΗ – pΔV

Which, if pressure is constant and gases can be considered as ideal can be transformed to:

ΔΕ = ΔΗ - RTLηρας

Where ∆n_gas 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:

ᎬᎫ AE = -5660kᎫ - 0.008314- , * 2Ꮽ8K * (-3moles) = -5653ᎬᎫ

Answer 2

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