 Ask Remember Me? electron Posts: 1, Reputation: 1 New Member #1 Aug 14, 2005, 12:40 PM
thermochemical equations
I have been given this thermochemical equation: and I need to calculate the quantity of heat liberated by the reaction of 50.0g of N2O(g) with excess NH3(g) and also the quantity of heat liberated by the reaction that produces 50.0g of N2(g)

Could anyone give me a clue as to how to go about working this out please?

Any help gratefully appreciated

Thanks x kp2171 Posts: 5,318, Reputation: 1612 Uber Member #2 Aug 17, 2005, 08:25 PM
I'll take a shot. Been a while since I've done this.

Well the equations says that when 2 moles of NH3 react with 3 moles of N2O you get 4 moles N2 and 3 moles H2O... with the reaction releasing 1010kJ.

So... first part is concerned with the amount of N2O used. Does it make sense that the amount of enegy exchanged depends on the amount of "stuff" used?

So if you use excess NH3 and a limited amount of N2O, then the N2O will determine how much enegy is released. It is the limiting reagent.

Use the molecular weight of N2O and the amount to determine how many moles of N2O were used. I'm not going to do this for you.

But lets say you come up with 6 moles N2O. You used twice the amount as in the equation. I believe, and this is where my brain is fuzzy, you simply say that its then 2 x 1010kJ.

Likewise, if you determined that 1.5 moles were used, 1.5/3 is the same as 1/2, or 0.5... so here you'd multiply 1010 x 0.5

Your number will probably not be as convenient as these. If the number is, say, 0.385 moles the rules are the same. 0.385/3 =?? And then take?? X 1010.

The second part is working backwards, but the same idea. 50.0g N2 is some number of moles. We see from the equation that 4 moles is what is made when 1010kJ is released. Sooooo... same idea. If only 1 mole N2 was made then 1/4 x 1010kJ is the answer.

Remember, many, many problems in general chemistry are answered by converting amounts to moles first and then using balanced equations to solve the problem at hand. If you cannot convert to moles from grams, you'll not pass.

Hope this helps. Hope its right. If not, ah well.

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