Here it says, the formation of ammonia is accomplished by the reaction between hydrogen and nitrogen gas. So here it says, calculate the enthalpy of reaction if the bond enthalpies of the N≡N, H—H, and N—H are 945, 432, and 391 kilojoules per mole respectively.
The steps we're going to take here is, we'll check to see if the chemical reaction is balanced. If it's not, then do the necessary steps to balance it. So here they're already drawn for us. It's already balanced for us. We have one mole of this. If Lewis structures are not given, then you will have to draw them as well. Luckily, it's been drawn for us.
Now step 1, for the reactants and products, multiply the coefficients of each bond type with its bond enthalpy value. Alright. So, if we take a look here, we have 1 N≡N bond and its energy is 945 kilojoules per mole. And here we have 3 H—H bonds, and each one we're told is 432. And then finally, we have how many N—H bonds? This one's tricky. We have 1, 2, 3 N—H bonds within NH3, but then it's times 2. So, 3 times 2 equals 6 total N—H bonds. So, it's going to be 6 N—H bonds here, each one is 391.
ΔH = 1 × 945 + 3 × 432 − 6 × 391 = -105 kJSo, what we need to realize here is that this will cancel out with the moles that we have and our answer will be in kilojoules. We're doing reactants minus products to find the ΔH of reaction. When we subtract these two, it gives me -105 kilojoules as the enthalpy of reaction when given those different bond enthalpies. So here, all it is is first knowing that you have a balanced equation, knowing that you've drawn your Lewis dot structures, and carefully examining the different types of bonds that exist. If you can do that, you'll be able to find the enthalpy of reaction for any of these given questions.
