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hey!

Well, from what I remember, this shouldn't be too hard. But I think you left out the enthalpy for breaking the NO2 bond...i looked it up and it should be around 33.2, but you should check your text for the right answer.

To solve this, simply substitute the the compund with its bond enthalpy. If the compound is on the left, the energy is positive, because you're breaking the bond. If its on the right, its negative, energy is always released during bond formation.

So here is what we have

NO2 = 33.2

NH3 = 45.9

H20 = -241.8

N2O = -X

Total H = -879.6

so now we simply set it up as an equation and solve it like its 9th grade algebra.

3(33.2) + 2(45.9) + 4(-X) + 3(-241.8) = -879.6

solve for X, you should get -86kj per more. But as I said earlier, check with your text for correct enthalpy amount.

Hope this helped!

delta H = delta H products - delta H reactants)

-879.6 = [4(0) +3(-241.8)] - [3(x) + 2(-45.9)]

-879.6 = -725.4 - 3X + 91.8

-879.6 + 724.4 - 91.8 = -3x

82.3 KJ / mol = x = NO2