What is the enthalpy for this reaction, given...?

[the_fella]

The reaction is 1. C(s)+ 1/2 O2 ---> CO They want us to find the enthalpy for this reaction.

Given:
2. C(s) + O2 ---> CO2 Delta H = -393.5 KJ/mole
3. CO + 1/2 O2 ---> CO2 Delta H = -283 KJ/mole

My instinct is to reverse equation 2, and then just add all three together. Everything cancels that way, and I get +110 KJ/mole, but the answer says it's -110 KJ/mole and that I should've reversed equation 3, but when I do that, something doesn't seem right:

C(s)+ 1/2 O2 ---> CO2
CO2 ---> CO + 1/2 O2

So we're left with C(s) ---> CO Is it ok that we're missing the half mole of oxygen here?

Was my mistake adding equation 1 to the other two? I thought we were supposed to do that.

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[sazerac]

You are supposed to add equation 2 to the reverse of equation 3. This produces equation 1 (which is what you wanted) and also tells you the enthalpy.

 

[NextStepTutor_1]

^ Yep.

This looks like:
2. C(s) + O2 ---> CO2 ∆H = -393.5 KJ/mole
3. CO + 1/2 O2 ---> CO2 ∆H = -283 KJ/mole

To get the equation C(s)+ 1/2 O2 ---> CO from these 2, we must flip the 3rd equation as such:
2. C(s) + O2 ---> CO2 ∆H = -393.5 KJ/mole
3. CO2 ---> CO + 1/2 O2 ∆H = +283 KJ/mole

Crossing out the CO2 and 1/2 O2 from products and reactants give us:
C(s)+ 1/2 O2 ---> CO
And so we just need to add the half enthalpies to get -110.5 kJ/mole

 

[the_fella]

^ Yep.

This looks like:
2. C(s) + O2 ---> CO2 ∆H = -393.5 KJ/mole
3. CO + 1/2 O2 ---> CO2 ∆H = -283 KJ/mole

To get the equation C(s)+ 1/2 O2 ---> CO from these 2, we must flip the 3rd equation as such:
2. C(s) + O2 ---> CO2 ∆H = -393.5 KJ/mole
3. CO2 ---> CO + 1/2 O2 ∆H = +283 KJ/mole

Crossing out the CO2 and 1/2 O2 from products and reactants give us:
C(s)+ 1/2 O2 ---> CO
And so we just need to add the half enthalpies to get -110.5 kJ/mole

Ooooh! I getcha!. 1/2 of the O2 cancels with the 1/2 O2, leaving 1/2 O2. I see what you did there! Thank you very much!