2CH 3 OH(l)+3O 2 (g) 4H 2 O(l)+2CO 2 (g) 2CH2OH (302 ( ) - ( 1 ) ( ) \Delta H=-1452.8 kJ/mol. What is the value of \Delta H if (a) the bc equation is multi plied throughout by 2

Question

Consider this reaction : 2CH 3 OH(l)+3O 2 (g) 4H 2 O(l)+2CO 2 (g) 2CH2OH (302 ( ) - ( 1 ) ( ) \Delta H=-1452.8 kJ/mol.  What is the value of \Delta H if (a) the equation is multi plied throughout by 2 , (b) the direction of the reac tion is reversed so that the products become the reactants and vice versa, (c) water vapor instead of liquid water is formed as the product?

Answer 1

Please consider a few adjustments to the answer provided by Dr. J.R.:

I believe that the "Final answer" should have a minus sign, as the -1452.8 overpowers the + 176.2.

Also, 2 lines up from that, the decimal .2 from -967.2 will cancel when essentially added to 1143.2, so the result will be 176.0, rather than 176.2. (The math at the end was apparently done using 176.0) So the "Final answer" should be -1276.8 kJ.

Now one more issue with the wording of the problem; this may be debatable, as very respectable people have used this form. The texts that I've used write it slightly differently. Feel free to offer your ideas.

The very first equation gives delta H in units of kJ/mol. However, my question about this is, per mol of what? When we speak of standard enthalpy of formation, yes, it's per mol of the substance formed. Or, with other terms such as heat of vaporization or fusion, it's per mol of the substance being vaporized or frozen.

However, when we write an equation for a general reaction, and delta H is given, that is the change for reacting 2 moles of methanol and 3 moles of oxygen gas to produce 4 moles water and 2 moles CO2. Therefore, IMHO, it makes sense to write delta H, but not delta H per mol. Then it follows that when doubling the coefficients, the delta H also doubles. If it really were a molar property, then the delta H per mole would remain constant, regardless of the number of moles involved.

Answer 2

2CH₃OH(l) + 3O₂(g) ==> 4H₂O(l) + 2CO₂(g) ... ∆H = -1452.8 kJ/mol

(a) if equation is multiplied by 2, ∆H is also multiplied by 2. ∆H = -2905.6 kJ/mole

(b) if equation is reversed, the sign of ∆H changes. ∆H = +1452.8 kJ/mole

(c) if H₂O(g) is formed instead of H₂O(l), we can find ∆H as follows:

2CH₃OH(l) + 3O₂(g) ==> 4H₂O(l) + 2CO₂(g) ... ∆H = -1458.2 kJ/mol

4H₂O(l) ==> 4H₂O(g) ... ∆H = ? but since we don't know this ∆H we need to find it (calculate it) from tabular values of ∆Hº_f of H₂O(l) and H₂O(g). So, ∆H for the conversion of 4 moles H₂O(l) to 4 moles of H₂O(g) is...

4 x ∆H_f H₂O(g) - 4 x ∆H_f H₂O(l)

4(-241.8 kJ/mol) - 4(-285.8 kJ/mol) = -967.2 - (-1143.2) = 176.0 kJ/mole

Now we have to add this to the original ∆H of -1452.8 kJ/mole

Final answer = -1452.8 + 176.2 = -1276.8 kJ/mole

I hope part (c) makes sense. In the original equation, water is a liquid. For part (c) it is a gas. So, we need to find the ∆H to convert liquid water to gaseous water and then add that to the ∆H of the original reaction.