formation question

[allthebest]

Consider the following information.

Compound	C6H6(l)	CO2(g)	H2O(l)
∆HfӨ / kJ mol–1	+49	+394	–286

                    C6H6(l) + 7.5O2(g) à 6CO2(g) + H2O(l)

Which expression gives the correct value of the standard enthalpy change of combustion for benzene (l), in kJ mol–1?

A.           12(-394) + (-286) - 2(49)

B.           12(394) + 6(286) - 2(- 49)

C.           6(-394) + 3(-286) - (- 49)

D.           6(394) + 3(286) - (- 49)

Why is the answer C?

why does the value for CO2 become minus while the sign of H2O stays the same? And where does the "3" come from?

Thanks

[Msj Chem]

The 3 should be the coefficient in front of the H₂O. Without it the equation is not balanced. 

And the enthalpy of formation of CO₂ is -394 kJmol^-1, not +394 kJmol^-1. 

Where did you get this question from?

Edited May 18, 2016 by Msj Chem

[fafamehr]

On 5/18/2016 at 12:25 PM, Msj Chem said:

The 3 should be the coefficient in front of the H₂O. Without it the equation is not balanced. 

And the enthalpy of formation of CO₂ is -394 kJmol^-1, not +394 kJmol^-1. 

Where did you get this question from?

https://sites.google.com/site/ibchemistryellesmerecollege/home/topic-15---energetics-thermochemistry-ahl/15-3-past-paper-questions 

It is on your website, question 12