7. Energy, Rate and Equilibrium
Bond Energy
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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 or bond enthalpies of the n triple bond n, h single bond h, and n single bond h r 945, 432, and 391 kilojoules per mole respectively. So the steps we're gonna take here is we're gonna check to see if the chemical reaction is balanced. If and if 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 more of this. If lewis structures is not given, then you will have to draw them as well. Luckily, I gave it to us drawn. 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 triple bond n bond and its energy is 945 kilojoules per mole. And here we have 3 h single bond 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 n h three, but then it's times 2. So 3 times 2 equals 6 total n h bonds. So it's gonna be 6 n h bonds here, each one is 391. So then, what we need to realize here is that this will cancel out with the moles that we have and our ants will be in kilojoules. So what we're gonna do here is 9 1 times 945, plus 3 times 432 gives us 2241 kilojoules. Here we're doing, reactants minus products to find the delta h of reaction. Then we have 6 times 391 which is 2346 kilojoules. When we subtract these 2, it gives me negative 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.
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