Asked by Nick
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
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
Answers
Answered by
bobpursley
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)= ...
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)= ...
Answered by
DrBob222
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.
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.
Answered by
Nick
Thanks for your help!
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