Energy, q, w, Enthalpy, and StoichiometrySource(s):

Consider the combustion of liquid methanol, CH3OH (l).

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. ?????

I think it is this but not sure.

+ 726.4 kJ

Now it makes sense with the H. The delta H for the doubled equation is 2*delta H for the original and the -1453 kJ should give you the correct answer.

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. ?????

I'm stuck but not as much as before. I keep looking for WORDS like combustion, equilibrium, etc but the "ans to 4 s.f." means they want a number. Perhaps the number is -726.5kJ/mol*2 = -1453 kJ. I'm still in the dark if this isn't right.

thank you that is the answer. -1453

It was driving crazy.

I did not * 2 just one simple step will mess ya up lol.. thanks for your help... your awesome.

My answer is +239

Finally achieved a gr8 success my ans is -239

To find the enthalpy change (ΔH) for a reaction, you can use the principle of Hess's Law. According to Hess's Law, the enthalpy change for a reaction can be determined by the difference in enthalpy between the products and the reactants.

In this case, the given reaction is the combustion of liquid methanol (CH3OH) where it reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The enthalpy change for this forward reaction is given as -726.5 kJ.

Part A:
To find the enthalpy change for the reverse reaction, you need to reverse the given reaction equation and the sign of the enthalpy change. Since the given reaction is already balanced, you can simply reverse the equation to get:

CO2 (g) + 2H2O(l) -> CH3OH (l) + 3/2O2 (g)

Since the reaction is reversed, the enthalpy change for the reverse reaction will have the opposite sign. Therefore, the enthalpy change for the reverse reaction is 726.5 kJ. This is the correct answer you provided.

Part B:
To balance the forward reaction with whole-number coefficients, you need to ensure that the number of atoms of each element is the same on both sides of the equation. Start by balancing the carbon (C) atoms, then the hydrogen (H) atoms, and finally the oxygen (O) atoms.

The balanced equation for the reaction is:

2CH3OH (l) + 3O2 (g) -> 2CO2 (g) + 4H2O(l)

This is the correct answer you provided.

Part C:
To calculate the enthalpy change (ΔH) for the balanced equation, you can use the coefficients of the balanced equation. For the balanced equation:

2CH3OH (l) + 3O2 (g) -> 2CO2 (g) + 4H2O(l) ΔH = -726.5 kJ

Since ΔH is an extensive property, it is directly proportional to the coefficients in the balanced equation. Therefore, the enthalpy change for the reaction represented by this equation is:

ΔH = (2 * -726.5 kJ) = -1453 kJ

So, the enthalpy change for the reaction is -1453 kJ.