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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Alycia Zemlak · Accepted Answer

a.)+726. 5 kJb.)2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(l), ΔH = − 1453.0 kJc.)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.

Alycia Zemlak · Answer

+756.5 kJb.) 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 H2OThe heat of reaction is twice as much. This is -1,453 kJc.) 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.

Johathan Wehner

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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