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What is the standard enthalpy of formation of liquid n-butanol, CH3CH2CH2CH2OH?

CH3CH2CH2CH2OH(l) + 6O2(g) → 4CO2(g) + 5H2O(l); ΔH° = -2675 kJ

Substance ΔH°f (kJ/mol)

CO2(g) -393.5

H2O(l) -285.8

Verified Answer :

MissPhiladelphia

For the answer to the question above on what is the standard enthalpy of formation of liquid n-butanol, CH3CH2CH2CH2OH if
CH3CH2CH2CH2OH(l) + 6O2(g) → 4CO2(g) + 5H2O(l); ΔH° = -2675 kJ
Substance ΔH°f (kJ/mol)
when the CO2(g) -393.5, H2O(l) -285.8

The answer is -328 kJ
I hope this helped you.

sebassandin

Answer:

\(\Delta _fH_{C_4H_{10}O}^0=-328kJ/mol\)

Explanation:

Hello,

Considering the specified chemical reaction, one writes the expression to compute this reaction's heat as shown below:

\(\Delta _rH^0=4\Delta _fH_{CO_2}^0+5\Delta _fH_{H_2O}^0-\Delta _fH_{C_4H_{10}O}^0\\\)

Now, solving for the standard enthalpy of formation of liquid butanol, one obtains:

\(\Delta _fH_{C_4H_{10}O}^0=4\Delta _fH_{CO_2}^0+5\Delta _fH_{H_2O}^0-\Delta _rH^0\\\Delta _fH_{C_4H_{10}O}^0=4(-393.5kJ/mol)+5(-285.8kJ/mol)-(-2675kJ/mol)\\\Delta _fH_{C_4H_{10}O}^0=-328kJ/mol\)

Best regards.

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