Question

The following equations show the oxidation of carbon and carbon monoxide to carbon dioxide.

C (s) + O2 (g) --> CO2 (g) Delta H = - x kJ mol-1

CO (g) + 1/2O2 (g) --> CO2 (g) Delta H = - y kJ mol-1

What is the enthalpy change, in kJ mol-1, for the oxidation of carbon to carbon monoxide?
C (s) + 1/2O2 (g) --> CO (g)

Choice of answers:
A. x + y
B. - x - y
C. y - x
D. x - y

I thought I understood this, but now I don't !

Answer 1

http://openstudy.com/study#/updates/521356b5e4b0450ed75da6ff
LOL i have already solved it

Answer 2

Yes but actually I don't understand exactly how you got to the answer! Could you please try and explain again?

Answer 3

have u read anything related to this or noo first tell me

Answer 4

I've studied Hess' law, combustion and formation energy cycles, is this question an example of one of these?

Answer 5

@Frostbite can u help him i am busy with some other work

Answer 6

Ok thank you!

Answer 7

I'll be happy to @chmvijay.

Yes this is just the application of Hess' law. It may be a little tricky in the start, until you get to know it.
Hess's law states that the standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which a reaction may be divided.

So we are going to divide the reaction and remember another relation:
For a reaction A -> B it then follow:
\[\Delta H ^{\Theta}(A \to B)=-\Delta H ^{\Theta}(B \to A)\]

Our final reaction is C (s) + 1/2O2 (g) --> CO (g), how would you combine the two given reactions to get out final one?

Answer 8

u arrange the first and second reaction of this in such way that u should get the thire reaction equation

Answer 9

since in ur reaction CO is at right hand side and even in second equation Co is at left hand side so u need to reverse it when u reverse the second equation the heat released -y becomes +Y

Answer 10

Ok thank you frostbite, I think that's the bit I don't understand how to do

Answer 11

If I reverse the second eqn and make it +y, I understand. But I don't understand why the 2 nd eqn is anymore related to the final eqn than before

Answer 12

Except,

so lets start with the first one:
C (s) + O2 (g) --> CO2 (g)

This look good so far, but we need to get rid of CO2 (g). We do so by adding the REVERSE 2. reaction:
CO2 (g) -> CO (g) + 1/2O2 (g)

add them together (reactants together and product together):
C (s) + O2 (g) + CO2 (g) -> CO2 (g) + CO (g) + 1/2O2 (g).

Reduce it:
C (s) + 1/2 O2 (g) -> + CO (g)
Exactly what we wanted.

So that mean hess law states the equation is going to be used like this:
\[\Delta H ^{\Theta}(1)+(-\Delta H ^{\Theta}(2))\]
It gets - because we used the reverse reaction.

Answer 13

Equation 1+ reverse equation 2 u add and cancel on opposite side if they are same like CO2 get cancelled like 1/2 O2 also get cancelled

Answer 14

then u get the required equation and the required energy released y-x

Answer 15

@Frostbite thank you i am sorry i interpreted in between Thank you for ur help too :)

Answer 16

I get it a bit more now! But how do you get rid of the half O2?

Answer 17

And thank you!

Answer 18

The one of the right

Answer 19

You have 1 O2 on the reactants and 1/2 on the products. So \(1-1/2=1/2\)

Answer 20

no need to apologize @chmvijay. You only contribute making it able to be seen from more views.

Answer 21

1 O2 first equation
1/2 on the products in second equation
if u cancel both then
1/2 O2 will be at first equation

Answer 22

Thank you very very much! I understand 100%! Thank you!

Answer 23

To both of you!

Answer 24

ur welcome dude :)