The combustion of carbon monoxide is represented by the equation below:

Question

The combustion of carbon monoxide is represented by the equation below:

CO (g) + 1/2O2 (g) ----> CO2 (g)

a) determine the value of the standard enthalpy change, for the combustion of CO at 298 K using the following info.

C (s) + 1/2 O2(g) ----> CO (g)
delta H at 298K = -110.5 kJ/mol

C(s) + O2 (g) ----> CO2
delta H = -393.5 kJ/mol

so far i have:
1/2 C (s) + 1/4 O2 (g) ----> 1/2 CO

and C (s) + O2 (g) ----> Co2

I'm kind confused as to what i'm supposed to do and how i'm supposed to use the info. am i on the right track? if not, could you tell me how it works? thanks

Answer 1

let A= CO2

B= CO

elements> A gives 393kj/mol heat
elements>B gives 110kj/mol
Question is B > A

elements >
B > A
You know elements >A
and elements > B, so
B>A must be the difference
-393 - (-110)= ...

Answer 2

Yes you are on the right track but you just need a little redirection.

Reverse C + 1/2O2 ==> CO and change sign of delta H.
Add C+O2 ==> CO2 and keep delta H as is.
Now add the two equations and see if you get CO + 1/2 O2 ==> CO2. Add delta Hs.
Thanks for posting your work.

Answer 3

Thanks for your help!

Answer 4

You're on the right track! To determine the value of the standard enthalpy change for the combustion of CO (g), you need to use the information given about the enthalpy changes for the reactions involving C (s), O2 (g), CO (g), and CO2 (g).

First, let's focus on the desired reaction:

CO (g) + 1/2 O2 (g) ----> CO2 (g)

You have correctly identified the reactions involving CO (g) and CO2 (g):

C (s) + 1/2 O2 (g) ----> CO (g)
ΔH at 298K = -110.5 kJ/mol

C (s) + O2 (g) ----> CO2 (g)
ΔH = -393.5 kJ/mol

Now, to determine the value of the standard enthalpy change for the desired reaction, you need to manipulate these equations and combine them to get the desired equation.

Since you want to cancel out CO, you can multiply the first equation by 1/2 to make it balanced:

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

Now, multiply the second equation by 1 to keep it balanced:

C (s) + O2 (g) ----> CO2 (g)

Now, add the two equations together:

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

Simplifying, we get:

3/2 C (s) + 3/2 O2 (g) ----> 1/2 CO + CO2 (g)

From this equation, you can see that the desired reaction will involve the combustion of 3/2 moles of C (s) and 3/2 moles of O2 (g).

Now, you need to determine the enthalpy change for this combined reaction. Since enthalpy is an extensive property, the ΔH values for the reactants and products need to be multiplied by the same factor as the stoichiometric coefficients.

Multiplying the first equation by 1/3, the second equation by 2/3, and adding them together, we can determine the ΔH for the combined reaction:

(1/3) * (-110.5 kJ/mol) + (2/3) * (-393.5 kJ/mol) = ΔH(combined)

Calculating this, the value of ΔH(combined) will give you the value of the standard enthalpy change for the combustion of CO at 298 K.