Consider the combustion of liquid methanol, CH3OH(l): CH3OH(l)+(3/2)O2(g) → CO2(g) + 2H2O(l); ΔH=−726.5kJWhat is the enthalpy change for the reverse reaction?

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blueotter787Lv1

11 Dec 2019

Consider the combustion of liquid methanol, CH₃OH(l): CH₃OH_(l)+(3/2)O_2(g) → CO_2(g) + 2H₂O_(l); ΔH=−726.5kJ
What is the enthalpy change for the reverse reaction?

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Coleen AmadoLv10

26 Nov 2020

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Explain why the answer is B...

Methanol is manufactured by the partial combustion of methane under pressure, using a copper catalyst:
CH₄(g) + Â½ O₂(g) â CH₃OH(l)
Enthalpies of combustion are (in kJ mol^-1):

CH₄(g), -881
CH₃OH(l), -726

Calculate the standard enthalpy change for the manufacture of 1 mole methanol by the reaction
CH₄(g) + Â½ O₂(g) â CH₃OH(l)

Follow the procedures based on Hessâs Law:
First write down the reactions corresponding to the enthalpies of combustion you have been given, reverse one of the equations remembering to change the sign of ÎH^o, then combine these equations to give the required process, and evaluate the associated ÎH^o
State the answer to three significant figures without decimal point, Â±xxx