Question

How would I solve this?

Answer 1

Give us your thoughts, dear.

Answer 2

I know you have to combine them to get the answer, but I remember learning that under certain circumstances, you change the signs. Unfortunately I don't remember what those circumstances are.

Answer 3

@DanJS
Hey there. Want a medal?

Answer 4

Remember that chart from yesterday ?

Answer 5

One second and I'll find it.

Answer 6

to get tho those products you need, it is a net enthalpy of all those reactions right?>

Answer 7

Yeah, but how do you know when the signs change?

Answer 8

i see now what is going on, it has been a few years since chem

Answer 9

http://www.chem.tamu.edu/class/majors/tutorialnotefiles/enthalpy.htm

Answer 10

look at example 2, it is just like this one

Answer 11

when you total up the reactions, you should end up with the given overall reaction... you may need to multiply some of the intermediate steps by a constant, and in tern the enthalpy for that step..

Answer 12

When you flip the reaction around the other way, you also flip the sign on the enthalpy

Answer 13

you want everything to cancel out on both sides of the net equation, so it leaves you with the overall reaction

Answer 14

I see how everything was canceled out.
So since nothing needs to be multiplied and nothing is flipped, I would just leave the enthalpy alone and combine them as is?

Answer 15

yeah , if you can cancel everything out without multiplying or flipping the equations around, then it is just the net of the given enthalpys

Answer 16

Okay so B) -304.1 kJ

Answer 17

yeah it looks like everything cancels as is , and leaves you with the Net equation without doing anything

Answer 18

sorry i am just learning this again now. lol

Answer 19

3/2 O2 ----> 3/2 O2

cancels

Answer 20

and one O3 from both sides

Answer 21

Leaves you with the net equation they give you ... so just add the enthalpies as is, and yeah -304 ..

Answer 22

but the rules are, if you flip the equation around, switch the sign on the enthalpy
and
if you need to multiply through by a constant, also multiply the enthalpy by that constant

Answer 23

I see. Thank you for your help! (you explain things a lot better than my teacher)

Answer 24

lol, i forgot all about this till i read the web page, i do recall doing it though in chem first semester college general chem