No, I'm still not in touch with you I don't think. Are you supposed to calculate dH1 and dH2 or are those given to you. I assume they have not been given and you must calculate them. If so then dHf certainly is relevant to the question. If you do dHrxn = (n*dHf products) - (n*dH reactants) you get dHrxn1. Then you do the same for the second reaction and that gives you dHrxn2. Then you combine the reverse of H1 rxn (and you reverse the sign whatever it happens to be)(see note below) and add it to the forward reaction for H2 (and keep the sign whatever it happens to be). That will give you the reaction you are looking for as well (add the reverse of rxn 1 and the forward rxn of 2 and see if that doesn't give you the reaction you want at the end). Then you add the NEW values with the NEW signs to give the dH for the NEW reaction. Any sign that is negative is exothermic for that reaction and any sign that is positive is endothermic for that reaction.
NOTE. Note that saying "reverse the sign" is the same as saying "change the sign to the negative" or "take the negative of that value" or any number of similar statements. I don't think I ever said anything about "changing the sign to a negative" but if I did I shouldn't have. In thermochemisty all kinds of equations are written and tabulated with delta H values. If the value is positive the reaction is endothermic and if negative the reaction is exothermic. If we don't like the equation as it is but we want the reverse of that equation, we simply reverse the equation and reverse the sign (a + sign becomes a - sign or a - sign becomes a + sign). Does that change whether it is exothermic or endothermic. You bet it does. If
A==> B is dH = -15 kJ then
B==> A is dH = +15 kJ and what was an exothermic reaction (-15 kJ) becomes an endothermic reaction (+15 kJ). I hope this helps but if not I don't mind trying again. If you pursue this further, please write the question first and comments later. Also, please avoid using all caps--that makes it difficult to read
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Homework Help: Chemistry
Posted by Anonymous on Thursday, November 14, 2013 at 8:45pm.
CONTINUE>>>>>>>>>>>
The enthalpy changes for two different hydrogenation reactions of C2H2 are:
C2H2+H2---->C2H4 Delta H 1 (there is a degree sign....standard enthalpy of formation??)
*****WAIT A SECOND, IF I USE THE HEAT OF FORMATION VALUES TO CALCULATE THEN WOULD THAT GIVE THE ENTHALPY OF RXN???...which is not relevant to this question,right???
*****But if you change the sign to a neg then how do you know that it is positive originally???I AM VERY VERY PERPLEXED,DR.BOB!!!!!!!!!
I DO UNDERSTAND THAT YOU HAVE TO SWITCH ACCORDING TO HESS'S Law, but I DO NOT UNDERSTAND HOW YOU KNOW THE SIGNS......!!!!!!!!!!!!!!!!!!!!
Why is it pos to neg ??????????
I thought it is the other way around!!!!
BUT DO YOU UNDERSTAND WHAT I DO NOT GET??? IF NOT, I WILL TRY TO CLARIFY IT.
THANK YOU FOR YOUR PATIENCE!!!!
C2H2+2H2---->C2H6 Delta H 2 (there is a degree symbol)
Which expression represents the enthalpy change for the reaction below?
C2H4+H2---->C2H6 Delta H = ?
Ignore the things below.....
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