Consider the combustion of liquid methanol (CH3OH (l)) :CH3OH (l) + 3/2 O2 (g) to CO2(g) + H2O(l) = – 726. 5 kJ?

Question

Could you please clarify the combustion reaction of liquid methanol (CH3OH)? The reaction can be represented as follows: 
[ 	ext{CH}_3	ext{OH} (l) + \frac{3}{2} 	ext{O}_2 (g) ightarrow 	ext{CO}_2 (g) + 	ext{H}_2	ext{O} (l) ]
This process releases an enthalpy of -726.5 kJ. What does this negative value indicate about the reaction's energy changes, and how is it relevant to the efficiency of using methanol as a fuel?

Dale Gaylord · Accepted Answer

a.)
+726. 5 kJ
b.)
2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(l), ΔH = − 1453.0 kJ
c.)
Since the forward reaction is exothermic it's more likely to be favored.
d.)
The proposed change in the equation would cause the transition H2O(g) → H2O(l) to be included in the equation's ΔH.  That transition is exothermic (heat is lost by the gas as it condenses) so the magnitude of ΔH would increase.

Anonymous · Answer

+756.5 kJ
b.) balance the forward reaction w whole number coefficients. What is delta H for tje reaction represented by this equation?  
To balance this equation, double all of the coefficients.
2 CH3OH + 3 O2 → 2CO2 + 2 H2O
The heat of reaction is twice as much. This is -1,453 kJ
c.) which is more thermodynamically favored?, the forward reaction or the reverse reaction?  
The negative sign on the enthalpy change means this much energy is produced by this reaction. To reverse this reaction, 1,453 kJ of energy would be required. One reason is that carbon dioxide and water are more stable than methanol. This is why the forward reaction is favored. 
d.) if the reaction were written to produce H2Og instead of H20l, would you expect the magnitude of delta H to increase, decrease, or at the same time?  
This does not make any sense. H2O and H2O1 are the same thing.

Anonymous

Consider the combustion of liquid methanol (CH3OH (l)) :CH3OH (l) + 3/2 O2 (g) to CO2(g) + H2O(l) = – 726. 5 kJ?

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