Calculate the standard enthalpy of formation ( in KJ/mol) of liquid methanol, CH3OH(1), using the following information: C(graph) + O2 CO2(g) AH° = -393.5 kJ/mol H2(8) + (1/2)02 → H20(1) AH° = -285.8 kJ/mol CH3OH(1) + (3/2)O2(3) → CO2(g) + 2H2O(1) AH° = -726.4 kJ/mol INPUT your answer HERE in 3 significant figures. Numerical answer only, exclude the units. Do not forget the +/ or - sign.

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Calculate the standard enthalpy of formation ( in KJ/mol) of liquid methanol, CH3OH(1), using the
following information:
C(graph) + O2 – CO2(g) AH° = -393.5 kJ/mol
H2(g) + (1/2)02 → H20(1) AH° = -285.8 kJ/mol
CH3OH(I) + (3/2)O2(g) → CO2(g) + 2H2O(1) AH° = -726.4 kJ/mol
INPUT your answer HERE in 3 significant figures. Numerical answer only, exclude the units. Do not forget the +/or - sign.
Transcribed Image Text:Calculate the standard enthalpy of formation ( in KJ/mol) of liquid methanol, CH3OH(1), using the following information: C(graph) + O2 – CO2(g) AH° = -393.5 kJ/mol H2(g) + (1/2)02 → H20(1) AH° = -285.8 kJ/mol CH3OH(I) + (3/2)O2(g) → CO2(g) + 2H2O(1) AH° = -726.4 kJ/mol INPUT your answer HERE in 3 significant figures. Numerical answer only, exclude the units. Do not forget the +/or - sign.
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