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Hess's Law question


Full Member
10+ Year Member
Jan 20, 2009
  1. Pre-Dental
    Using Hess's Law, what is the ▲H rxn for:

    2C(s) + O2(g) >>>>> 2CO(g)

    C(s) + O2 >>>>> CO2(g) ▲H= -393.7KJ/mol
    CO(g) + ½O2 >>> CO2(g) ▲H= -283.3KJ/mol

    When I did this problem I got an answer that was way off apparently. By using ▲H = Σ▲H(products) - Σ▲H(rxtnts) Shouldn't you write(?):
    -((2(-283)) - 2(-393)? This would give me +566 -(-786) = +1352KJ. I was really confused to see that the answer was -220KJ and the way that it was set up seemed opposite to me. The answer said 2(-393KJ)-(-2(283KJ)) = -220KJ. To me it looks like the ▲H = reactants - products. Anyways, for this particular problem there's something i'm not seeing... please help!! Thanks


    Full Member
    10+ Year Member
    Mar 2, 2009
    1. Pre-Dental
      First of all, Hess's Law states that enthalpies of reactions are additive. Therefore, after you manipulate the reactions to look like the reaction that you want, you add them, not subtract.

      So, the first reaction should be multiplied by 2 and the second should be flipped (multiply by -1) and multiplied by 2. When you do this to both equations you will see that the 2CO2's will cancel out and 2O2 and O2 will cancel to become 1O2 on reactant side. You will also still have 2C on the reactant side to give you a net reaction of:
      2C + O2 --> 2CO which is what you want.

      So, now adding what you manipulated you get:
      (2(-393.7)) + (2(283.3)) = -220.8 kJ

      Hope this helps.
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