Question

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

Answer 1

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

Step-by-step 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.