# Solve Chem Q2: ΔH° for Combustion & NaOH Formation

• physicsman2
In summary: I think I'm getting it now. Yes, the first reaction starts with 1 mole of methanol, and the amount of heat evolved is per mole of methanol.
physicsman2

## Homework Statement

1)The complete combustion of 1.47 g of methanol produces 29.3 kJ of heat. Determine the ΔH° for the reaction and its sign.
CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(g)

2) The value of ΔH° for the reaction below is -126 kJ. How much energy is released when 2.00 mol of NaOH is formed in the reaction?
2Na2O2(s) + 2H2O(l) → 4NaOH(s) + O2(g)

## Homework Equations

Dimensional Analysis

## The Attempt at a Solution

1) I know the answer is -638 kJ but I don't know how you can get that answer. I know why it's negative because heat is produced. I think that you first find the molar mass by dividing 1.47g by 32(the molar mass of methanol) and then multiplying that by 29.3, but the units don't make sense. Thanks for any help.

2)I know how to get the answer, but can anyone explain how and why the answer is the answer? Multiply 2 mol NaOH by -(-126kJ/4mol NaOh) to get the answer.

physicsman2 said:
I think that you first find the molar mass by dividing 1.47g by 32(the molar mass of methanol) and then multiplying that by 29.3, but the units don't make sense.

Find the molar mass by dividing sample mass by molar mass? Show what you are doing and what units you are getting.

I know how to get the answer, but can anyone explain how and why the answer is the answer? Multiply 2 mol NaOH by -(-126kJ/4mol NaOh) to get the answer.

That's not correct. Once again, show what units you get.

--

1)1.47g methanol x 1 mol/32g methanol x 29.3 kJ/1 mol, is this how to do this

2)2 mol NaOH x -(-126kJ/4 mol NaOH) <--- 4 mol from reaction

physicsman2 said:
1)1.47g methanol x 1 mol/32g methanol x 29.3 kJ/1 mol, is this how to do this

I hate to repeat questions, but let's try again. What units did you get?

2)2 mol NaOH x -(-126kJ/4 mol NaOH) <--- 4 mol from reaction

Sorry, I have missed 4 mol NaOH in your previus post. That's OK. Just multiply and cancel.

--

1) sorry i thought you wanted me to do the problem. The way i did it I get kJ but that's only because i had 29.3kJ/1 mol. I have no idea why it would be 1 mol if it's supposed to be there.

2) could you explain that's the way to do the problem. I know how to do it but i have no idea why?

Thanks a lot

physicsman2 said:
1) sorry i thought you wanted me to do the problem. The way i did it I get kJ but that's only because i had 29.3kJ/1 mol. I have no idea why it would be 1 mol if it's supposed to be there.

kJ is wrong. You need to express the result in kJ/1 mole of methanol.

2) could you explain that's the way to do the problem. I know how to do it but i have no idea why?

Just watch your units. You know there are 126kJ/4mol NaOH (let's ignore sign for clarity). You know you have 2 moles of NaOH. When you multiply you get

$$2 [moles\ of\ NaOH] \frac {126 [kJ]} {4 [moles\ of\ NaOH]}$$

Moles of NaOH cancel out and you are left with

$$2 \frac {126 [kJ]} 4 = 63 kJ$$

Units are OK, so the result must be OK.

--

i'm fine on the second one, just that the negative goes away because energy is released right?

on the first one, the reason I ask about the units because the choices given were all in kJ, which is why I was confused on it, the answer was -638 kJ

physicsman2 said:
i'm fine on the second one, just that the negative goes away because energy is released right?

Yes.

on the first one, the reason I ask about the units because the choices given were all in kJ, which is why I was confused on it, the answer was -638 kJ

So the answers are not entirely correct, or what they mean is -638 kJ per "reaction as written" - but in this case that's the same as "per mole of methanol".

Note, that in the second reaction amount of heat is reported per 4 moles of NaOH - or per mole of O2, but it can be also per 2 moles of Na2O2 or 2 moles of water - this is all in one reaction. Sometimes it is important how the reaction is written:

Na2O2 + H2O -> 2NaOH + 1/2O2

would be correct, with ΔH° of -63 kJ. So the first reaction starts with one mole of methanol, and amount of heat evolved is per mole of methanol.

--

thanks so much for all your help

## 1. What is the formula for calculating ΔH° for combustion?

The formula for calculating ΔH° for combustion is ΔH° = Σ(ΔH°f(products)) - Σ(ΔH°f(reactants)), where ΔH°f represents the standard enthalpy of formation.

## 2. How do you determine the standard enthalpy of formation for a compound?

The standard enthalpy of formation for a compound can be determined by measuring the heat released or absorbed during a reaction and using the given standard enthalpies of formation for the reactants and products in the above formula.

## 3. What is the significance of ΔH° for combustion?

ΔH° for combustion is a measure of the overall energy change that occurs during the combustion of a compound. It can provide information about the stability of the compound and its ability to release energy in the form of heat.

## 4. How is ΔH° for NaOH formation different from ΔH° for combustion?

ΔH° for NaOH formation is specifically focused on the formation of the compound NaOH, while ΔH° for combustion can refer to the energy change in any combustion reaction. Additionally, ΔH° for NaOH formation only considers the reactants and products involved in the formation of NaOH, whereas ΔH° for combustion takes into account all reactants and products in the reaction.

## 5. Can the value of ΔH° for combustion or NaOH formation vary?

Yes, the value of ΔH° can vary depending on the conditions under which the reaction takes place. Standard conditions, such as a specific temperature and pressure, are typically used to calculate standard enthalpies of formation. However, if the reaction occurs under different conditions, the values may differ. It is important to specify the conditions when reporting or using ΔH° values.

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