Asked by Anonymous
The equilibrium constant for the reaction 2NO(g) + O2(g) <--> 2NO2(g) is Kp=1.48x10^4 at 184C. Calculate Kp for 2NO2(g) <--> 2NO(g) + O2(g).
I know the equation is Kp=Kc(RT)x Delta n
Kp is what we're solving for
Kc= I can't seem to find.
R=.0821
T=184+273
Delta n= 1
I was wondering if I had set this up right and where I go from here because I'm stuck at this point.
I know the equation is Kp=Kc(RT)x Delta n
Kp is what we're solving for
Kc= I can't seem to find.
R=.0821
T=184+273
Delta n= 1
I was wondering if I had set this up right and where I go from here because I'm stuck at this point.
Answers
Answered by
DrBob222
I think you are missing the boat.
Kp for 2NO + O2 ==> 2NO2 Kp = 1.48E3
You want Kp for
2NO2 ==> 2NO which is just the reverse of the original equation. For that the new Kp = 1/old Kp (i.e., just the reciprocal of old Kp).
Kp for 2NO + O2 ==> 2NO2 Kp = 1.48E3
You want Kp for
2NO2 ==> 2NO which is just the reverse of the original equation. For that the new Kp = 1/old Kp (i.e., just the reciprocal of old Kp).
Answered by
Anonymous
Oh wow. I really need to stop over thinking this stuff, I tend to do that a lot. Thank you very much.
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