Calculate the Delta H for the reaction:
NO(g) + O(g) --->NO2(g)
given the following information:
NO(g) + O3(g) ---> NO2(g) +O2(g) DH =-198.9 kJ
O3(g) --->3/2 O2(g) DH= -142.3 kJ
O2(g) --->2O(g) DH = 495.0kJ
I know that the answer is -304.1 kJ. I somehow managed to get it when I switched the second equation around, making 142.3 kJ and switched the third, making it -495.0. I also multiplied -495.0 by 1/2, giving me -247.5... I got the answer. But I want to know why and how? Why did it work when I multiplied it by 1/2?
Thanks!
NO(g) + O(g) --->NO2(g)
given the following information:
NO(g) + O3(g) ---> NO2(g) +O2(g) DH =-198.9 kJ
O3(g) --->3/2 O2(g) DH= -142.3 kJ
O2(g) --->2O(g) DH = 495.0kJ
I know that the answer is -304.1 kJ. I somehow managed to get it when I switched the second equation around, making 142.3 kJ and switched the third, making it -495.0. I also multiplied -495.0 by 1/2, giving me -247.5... I got the answer. But I want to know why and how? Why did it work when I multiplied it by 1/2?
Thanks!
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So Hess' law pretty much states that when you add up reactions to get a final reaction, you can find the delta H of the final reaction by adding up the delta H's of all the other reactions.
To do so, you must cancel out all unnecessary molecules that are not in the final reaction.
The first given reaction can stay the same because it has an NO(g) on the left and an NO2(g) on the right which we also have in the final reaction. So that stays the same.
The second given reaction though must be reversed. We know this because there is no O3 in the final reaction. We have to cancel it out somehow. In the first given reaction, there is an O3(g) on the left side. To cancel it out, we need to find an O3(g) on the right side. By reversing the second given reaction, we will have an O3(g) to cancel the other one out. However, like you did, you must change the sign of the delta H when reversing a reaction.
The third given reaction is most tricky. We somehow need to get rid of O2(g) because it is not in the final reaction but appears in all of the given reactions. We also need to come up with just one O(g) molecule on the left side of the reaction for the final reaction. Right now we have 2O(g) and on the right side. By reversing the reaction and multiplying everything by 1/2, we are able to cancel out the O(g) to get just 1O(g) on the left. And also, by reversing and multiplying by 1/2, we also find that it will cancel out all the O2(g) as well.
Now we just add up the new Delta H values and find it to be the right answer.
GOOD LUCK!!!
To do so, you must cancel out all unnecessary molecules that are not in the final reaction.
The first given reaction can stay the same because it has an NO(g) on the left and an NO2(g) on the right which we also have in the final reaction. So that stays the same.
The second given reaction though must be reversed. We know this because there is no O3 in the final reaction. We have to cancel it out somehow. In the first given reaction, there is an O3(g) on the left side. To cancel it out, we need to find an O3(g) on the right side. By reversing the second given reaction, we will have an O3(g) to cancel the other one out. However, like you did, you must change the sign of the delta H when reversing a reaction.
The third given reaction is most tricky. We somehow need to get rid of O2(g) because it is not in the final reaction but appears in all of the given reactions. We also need to come up with just one O(g) molecule on the left side of the reaction for the final reaction. Right now we have 2O(g) and on the right side. By reversing the reaction and multiplying everything by 1/2, we are able to cancel out the O(g) to get just 1O(g) on the left. And also, by reversing and multiplying by 1/2, we also find that it will cancel out all the O2(g) as well.
Now we just add up the new Delta H values and find it to be the right answer.
GOOD LUCK!!!