Question

Help please? http://oi64.tinypic.com/19sb36.jpg

Answer 1

Which equation will you have to reverse so that the reactants will be the same as in the equation you have to solve for?

Answer 2

Would it be the 3rd one?

Answer 3

Since there is an O2 at the end

Answer 4

one you definitely need to reverse, the third reaction though is a bit strange

\[O_{2g} \rightarrow 2O_{g}\]

Answer 5

Let's first work with the first 2 reactions.

Answer 6

\[NO_{2g} + O_{2} \rightarrow NO_{g} + O_{3g} \]

Answer 7

Why did I do this? and what happens to the enthalpy sign

Answer 8

The 3rd one is throwing me off because 2O is not in the equation you need to solve for, also the 2O(g) can't be cancelled out.

The equation you need to solve for has 2NO2 on the reactant side and 2NO on the product side, so one reaction needs to be reversed.

Answer 9

You want certain ones to cancel out and to have products and reactants where they need to be for the last one. When you reverse it you change the sign of the delta heat.

Answer 10

@photon336 The enthalpy becomes positive

Answer 11

\[2(O_{3g} \rightarrow \frac{ 3 }{ 2 }O_{2} , (\Delta H = -142 \frac{ kj }{ mol })) = 2O_{3} \rightarrow 3O_{2}, \Delta H -284 \frac{ kj }{ mol }\]

Answer 12

Where did the -284 come from?

Answer 13

Oh. Never mind. I see now.

Answer 14

photon336 multiplied the 2nd equation as well. so it is -142.3*2

Answer 15

I see. Thank you

Answer 16

\[2NO_{2} + 2O_{2} \rightarrow 2NO + 2O_{3}, \Delta H, +396 \frac{ kj }{ mol }\]

Answer 17

hey @staldk3 do you get why @LeibyStrauss and I flipped the first reaction and multiplied everything by 2?>

Answer 18

You multiplied so that you got rid of the fraction.

Answer 19

yep, and why did I flip the first one?

Answer 20

So the reactants/products will match the same sides as the one we are solving for.

Answer 21

cool, now last step we just need to see what happens when we combine both of the reactions that we did before

Answer 22

\[2O_{3} \rightarrow 3O_{2}\]

\[2NO_{2} +2O_{2} \rightarrow 2NO + 2O_{3}\]

So we have

\[2NO_{2} \rightarrow 2NO + O_{2}\]

Answer 23

if you notice that O3 are in the reactants and product side they cancel out

Answer 24

I was just looking at that actually.

Answer 25

@staldk3 to find the total enthalpy what do you think we need to do?

Answer 26

Why isn't the 3 carried over?

Answer 27

add them all together and change the signs of the ones that got reversed.

Answer 28

we have 3 moles of O2 on the product side and 2 moles on the reactant side

Answer 29

so we're left with one mole on the product side of O2

Answer 30

Here's what we got from the first two reactions. we need to add these up in the end.

but ask yourself how did I come to get these numbers?

\[\Delta H_{1} = -284 \frac{ kj }{ mol}\]
\[\Delta H_{2} = +396 \frac{ kj }{ mol }\]

Answer 31

The first one you got it by multiply by 2

Answer 32

yep but we also flipped the reaction too. so when we do that we change the sign of the enthalpy.

Answer 33

what about the second one how did I get +396?

Answer 34

Unsure about that one

Answer 35

Oh you did the same. You multiplied the -198 by 2

Answer 36

yes

Answer 37

so the goal of this now is we can combine both these two reactions to get the final answer

Answer 38

and we always add up the enthalpies from each step to get the total entlahpy of reaction

Answer 39

112

Answer 40

you get this whole thing right?

Answer 41

I'm getting it better. I don't really understand the point of all of it

Answer 42

think of it this way. You need to figure out the enthalpy of reaction for the total reaction and your given two separate reactions. You need to reverse the reactions as necessary and balance them. and then add them up to get the final equation when needed.

Answer 43

Ok

Answer 44

Take a look at this and show me what you see. this is everything I did for this problem.


2(O3g→32O2,(ΔH=−142kjmol))=2O3→3O2,ΔH−284kjmol



2NO2+2O2→2NO+2O3,ΔH,+396kjmol


Then I combined both of these reactions


2O3→3O2


2NO2+2O2→2NO+2O3


So we have this:

Answer 45

I get the canceling out. So to pick the one that needs to be reversed it needs to be able to match what you're looking for?

Answer 46

yep

Answer 47

Can you ever have more than one to reverse?

Answer 48

it depends on the reactions you're given. sometimes you'll just have to reverse one sometimes both. but when you do this both reactions must give you the reaction they are asking you for.

Answer 49

And you still change the sign of the delta heat?

Answer 50

whenever you change the direction of the reaction you change delta H too

Answer 51

Alright. Thanks

Answer 52

no problem