Consider the combustion of liquid methanol,
CH3OH (l)
CH3OH (l) +3/2O2 (g) = CO2 (g) + 2H2O(l) H = -726.5 kJ
What is for the reaction represented by this equation?
Express your answer using four significant figures
I saw your post earlier and didn't understand it; therefore, I passed it by. I assume no one else understood it either since you've received no responses. Perhaps you could rephrase the question because it makes no sense to me as is.
CH3OH (l) +3/2O2 (g) = CO2 (g) + 2H2O(l) H = -726.5 kJ
Part A:
What is the enthalpy change for the reverse reaction?
Express your answer using four significant figures.
Answer I got right: 726.5 kJ
Part B:
Balance the forward reaction with whole-number coefficients:
CH3OH (l) +3/2O2 (g) = CO2 (g) + 2H2O(l) H
Answer I got right: 2,3,2,4
Part C:
What is for the reaction represented by this equation?
Express your answer using four significant figures.
Answer : I am stuck. ?????
Energy, q, w, Enthalpy, and StoichiometrySource(s):
This is the subject that I am on.
Opps I left off what is H
Part C:
What is H for the reaction represented by this equation?
Express your answer using four significant figures.
Answer : I am stuck. ?????
To determine the standard enthalpy change (ΔH) for the reaction represented by the given equation, we can simply use the value provided in the equation:
ΔH = -726.5 kJ
The negative sign indicates that the reaction is exothermic, as it releases energy in the form of heat.
Therefore, the value for the reaction enthalpy is -726.5 kJ.