Question

Given the enthalpies of reaction below, determine the enthalpy change for the reaction:

P4O6 (s) + 2O2 (g) → P4O10 (s)


P4 (s) + 3O2 (g) → P4O6 (s)

∆H = -1640.1 kJ
P4 (s) + 5O2 (g) → P4O10 (s)

∆H = -2940.1 kJ
Answer

-4580.2 kJ

-619.8 kJ

-1300.0 kJ

+4580.2 kJ

Answer 1

You basically need to put the 2 reactions together in a way that gives you the top reaction.

Answer 2

You can write the reaction in reverse if you change the sign of the reaction enthalpy.

Answer 3

thats the thing, i really need help on this one, unlike the othertwo i dont know where to start

Answer 4

Ok. This one is a bit tougher, so let's go slow. :)

Answer 5

yes please

Answer 6

I mentioned that you can write a reaction in reverse if you change the sign of the reaction enthalpy. I'll show you what I mean.

Answer 7

thanks

Answer 8

P4O6 -> P4 + 3O2 ΔH=1640.1 kJ

Answer 9

Do you see how I did that?

Answer 10

so far so good

Answer 11

Ok.

Answer 12

Let's add the reactants and products of the two reactions together and see what we get.

Answer 13

alright

Answer 14

P4O6 + P4 + 5O2
These are the reactants.

Answer 15

for the first problem?

Answer 16

Adding the second reaction's reactants and the first reaction's reactants after I flipped it around.

Answer 17

oh gotcha

Answer 18

Make sense?

Answer 19

yes

Answer 20

Hold on, let me write them both out.

Answer 21

alright i have time haha

Answer 22

:)
P4O6 -> P4 + 3O2
P4 + 5O2 -> P4O10

Answer 23

These are the reactions I am adding together.

Answer 24

im with you so far

Answer 25

:)

Answer 26

Now for the products.
P4O10 + P4 + 3O2

Answer 27

So the whole reaction is
P4O6 + 5O2 + P4->P4O10 + P4 + 3O2

Answer 28

And anything that appears both in the reactants and products isn't part of the reaction. They are basically just spectating, so we can leave them out.

Answer 29

alright so now what

Answer 30

Are you following me?

Answer 31

struggling but suprisingly yes

Answer 32

:)

Answer 33

Can you write the reaction without showing compounds in both the reactants and products?

Answer 34

not for real :/

Answer 35

Ok, it 's alright. I'll show you. :)

Answer 36

thank you haha

Answer 37

P4O6 + 2O2 ->P4O10

Answer 38

:) Do you see how I did that?

Answer 39

The. P4's just cancel out completely.

Answer 40

But only some of the oxygens cancel.

Answer 41

yeah i see it now, im catching on i think

Answer 42

Cool! This is the same reaction as the original question.

Answer 43

So to find the enthalpy, we add the enthalpies of the reactions in the same way I added the reactions.

Answer 44

The first reaction 's enthalpy (after I flipped it ) + the second reaction's enthalpy.

Answer 45

Can you do that part?

Answer 46

sorta, could you show me for the sake of me being a visual learner? lol

Answer 47

Ok.
1640.1 + (-2940.1)=

Answer 48

I think that comes to -1300

Answer 49

yeah thats right. i got that now, what would i do with out you haha, thanks so much

Answer 50

Haha. :) You are very welcome.