I really don't know what you're problem is but let me start from the beginning. I think if you had tried following (perhaps you did) those posts you could have explained your problem a little more clearly. But here goes. All do NOT start out negative. Some of those dHf values are + and some are -. You know what they are when you look them up in the table. You don't know if a reaction is negative or positive until you do the dHrxn = (n*dHproducts)-(n*dH reactants) thing. That gives you a dHrxn which will be one or the other. Have you tried doing them?
I looked up dHf for C2H2(g) in my text and it is listed as 226.7 kJ/mol
C2H4(g) is 52.26 kJ/mol
C2H6(g) is -84.86 kJ/mol
H2(g) of course is zero as are all free elements. See two of them are + and one is -.
So for C2H2 +H2 ==> C2H4
dHrxn is (n*dHproducts) - (n*dH reactants).
dHrxn = (1*+226.7) - [(1*52.26) +1*0]
dHrxn = (+226.7)-(52.26) = +174.44 so this first one we've done is an endothermic reaction because dHrxn is +. Remember from the other post we reverse the first reaction and changed the sign so I will write it now as
C2H4 ==> C2H2 + H2 dH = -174.44 kJ/mol
(You see that if the forward reaction is +174.44 then the reverse reaction is -174.44. That's the way it works.
#2 reaction is
C2H2 + 2H2 ==> C2H6
Following what we did above this one is
dHrxn = (1*-84.86) - [(1*226.7)+ (2*0)]
dHrxn = -84.86 - 226.7 = -311.56 kJ/mol so that second reaction is exothermic since it is negative. Now if we add these two we get this:
C2H4 ==> C2H2 + H2 dH = -174.44
C2H2 + 2H2 ==> C2H6 dH = -311.56
C2H4 + C2H2 + 2H2 ==> C2H2 + H2 + C2H6
Now cancel common molecules on each side and you see C2H2 cancels. One mole H2 cancels. We are left with
C2H4 + H2 ==> C2H6 which is the equation you wanted in the problem and since we added the two equations to get the final equation we now add the two dH values to find dH reaction. That is -174.44 + (-311.56) = -486.00 kJ/mol.
So reaction 1 as written when we added is -174.44 and it by itself is exothermic. The second reaction as written is -311.56 which is exothermic and the total reaction you get from putting these two together is -486.00 kJ which is exothermic. In this case both reactions are exothermic and the total is exothermic BUT both could be positive, both negative, or one each to give a final equation that is either + or - (or zero) however it turns out.
Back to your other question about why did I just switch it to negative? The answer is because I REVERSED the equation. When equations are reversed the dH for the reversed equation is ALWAYS the negative of the forward direction. So if the forward reaction is + the reversed equation is - and if the forward rxn is - the reversed rxn will be +. Why did I reverse it in the first place? Because that's the ONLY way you can get equation 1 and equation 2 to add together to get the final equation. Try adding eqn 1 and eqn 2 as is without reversed #1 and you get this
C2H2 + H2 ==> C2H4
C2H2 + 2H2 ==> C2H6
2C2H2 + 3H2 ==> C2H4 + C2H6 and that AIN'T the equation you're looking for. right? BUT if you reverse #1 and add in #2 as is you get the equation you want. You probably already understand this since you said earlier that you understood Hess' Law. I hope this helps. That's a step by step, number by number solution. I've proofed this so I don't think I have any typos or omissions.
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