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.
It appears from the answer that I'm the one that answered but your last question really is too confusing to me to make sense of it.
Let me point out that it is never a good idea to break a question up into two or more parts and make two or more post of it. What you are experiencing is a good example of that.
For this question my answer stands. If you reverse dH1 (and the equation that goes with it) to make it -dH1 and add the two equations, you get -dH1 + dH2 = the expression you want and that would be rearranged to make it dH2-dH1. That makes the answer to this question C. I have no idea how this relates to some other question you may have posted recently. If you wish you might try posting ALL of the question and ALL of the possible answers AND YOUR WORK and let us take a look at it. As to exo or endothermicity, I can't even guess without knowing values for dH1 and dH2.
To clarify the question, you are asking about the change in enthalpy (ΔH) for the reaction C2H4 + H2 -> C2H6, given the enthalpy changes for the hydrogenation reactions of C2H2. The enthalpy changes for the two different hydrogenation reactions of C2H2 are:
1. C2H2 + H2 -> C2H4 ΔH1
2. C2H2 + 2H2 -> C2H6 ΔH2
You are confused because the calculation of the enthalpy of formation showed that the change in enthalpy (ΔH1) for the reaction in question is negative. However, after switching the ΔH1, you would expect it to be positive, based on your understanding of exothermic reactions, which have a negative enthalpy (ΔH).
To find the enthalpy change for the reaction C2H4 + H2 -> C2H6, you need to reverse the ΔH1 (change its sign to -ΔH1) and then add it to the ΔH2.
So, the correct expression representing the enthalpy change for the reaction is:
-ΔH1 + ΔH2
Now, you can select the corresponding option from the provided choices:
A. ΔH1 + ΔH2
B. ΔH1 - ΔH2
C. ΔH2 - ΔH1
D. -ΔH1 - ΔH2
The correct answer is D. -ΔH1 - ΔH2, as explained above.