## 4D.15

AlyssaYeh_1B
Posts: 100
Joined: Sat Aug 17, 2019 12:16 am

### 4D.15

From the data the question gives, I did the calculation: -1500 - (-1300 + 2(-286)) = 312kJ/mol. However, the answer key shows that it's -312kJ/mol. What did I do wrong?

Eugene Chung 3F
Posts: 142
Joined: Wed Nov 15, 2017 3:03 am

### Re: 4D.15

AlyssaYeh_1B wrote:From the data the question gives, I did the calculation: -1500 - (-1300 + 2(-286)) = 312kJ/mol. However, the answer key shows that it's -312kJ/mol. What did I do wrong?

If you add the standard enthalpies up, it should be -1300 +1560 +2(-286) which comes out to -312

Jonathan Gong 2H
Posts: 105
Joined: Sat Jul 20, 2019 12:16 am

### Re: 4D.15

I think that the enthalpy given in the problem is enthalpy of combustion. Meaning that the enthalpy of formation of each respective molecule would be -enthalpy of combustion. Thus, when you do your calculations, it would be dHrxn = 1560 kJ/mol - 1300 kJ/mol - 2(286 kJ/mol). Which equals -312 kJ/mol.

Return to “Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)”

### Who is online

Users browsing this forum: No registered users and 1 guest