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Nostrana [21]
3 years ago
11

Ethanol boils at 78.4 °C with \DeltaΔHvap = 38.6 kJ/mol. A 0.200-mol sample of ethanol is heated from some colder temperature up

to 78.4 °C, which requires 1.05 kJ of heat, and then vaporized. What will be the total amount of heat required (for both the heating and the vaporizing)?
Chemistry
1 answer:
NNADVOKAT [17]3 years ago
4 0

Answer:

Q_T=8.77kJ

Explanation:

Hello,

In this case, the total heat is computed via:

Q_T=n*\Delta H_{vap}+1.05 kJ\\

It means, by adding all the involved heat, which simply result:

Q_T=0.200mol*38.6 kJ/mol+1.05 kJ\\Q_T=8.77kJ

Thus, such heat must be provided to the ethanol, that is why it remains positive.

Best regards.

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ΔGºrxn=   202.5 kJ + 0 kJ - 119.7 kJ  = 82.8 kJ

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0 = 116 kJ -T (0.113 kJ/K)

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ΔG = -RTlnK and solve for K

- ΔG / RT = lnK  ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

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