Question

I know I posted this question before, but can you clarify it??

MY QUESTION IS AFTER YOU SWITCHED THE DELTA H1 THE CHANGE IN THE ENTHALPY IS NEGATIVE.... BUT THAT DOES NOT MAKE SENSE BECAUSE THE CHANGE IN ENTHALPY OF THE ORIGINAL DELTA H1 IS NEGATIVE(ACCORDING TO THE CALCULATION OF THE ENTHALPY OF FORMATION YOU TOLD ME TO DO EARLIER...) AND SO AFTER SWITCHING IT WOULDN'T IT BE POSITIVE?????AND PLUS THIS IS PUTING THINGS TOGETHER WHICH IS EXOTHERMIC WHICH HAS A NEGATIVE H......PLEASE EXPLAIN..........


The enthalpy changes for two different hydrogenation reactions of C2H2 are:

C2H2+H2---->C2H4 Delta H 1

C2H2+2H2---->C2H6 Delta H 2


Which expression represents the enthalpy change for the reaction below?

C2H4+H2---->C2H6 Delta H = ?



A. Delta H 1 + Delta H 2
B. Delta H 1 - Delta H 2
C. Delta H 2 - Delta H 1
D. -Delta H 1- Delta H 2


You said....
Reverse H1 (make - dH) and add to H2.

Answer 1

See your post above.