The equilibrium constant, Kc, for the following reaction is 5.19×10-3 at 286 K. 2NOBr(g) <-->2NO(g) +...
The equilibrium constant, Kc, for the following
reaction is 5.19×10-3 at
286 K.
2NOBr(g)
<-->2NO(g) +
Br2(g)
Calculate Kc at this temperature for the following
reaction:
NO(g) +
1/2Br2(g)
<-->NOBr(g)
Kc=?
Homework Answers
The given reaction is:
2 NOBr(g) <-> 2 NO(g) + Br2(g)
Kc = [NO]^2[Br2] / [NOBr]^2
= 5.19*10^-3
The required reaction is:
2 NO(g) + Br2(g) <-> 2 NOBr(g)
Use:
Kc’ = [NOBr] / [NO][Br2]^1/2
= (1/Kc)^1/2
= (1/5.19*10^-3)^1/2
= (192.7)^1/2
= 13.9
Answer: 13.9
Add Answer to:
The equilibrium constant, Kc, for the following
reaction is 5.19×10-3 at
286 K.
2NOBr(g)
<-->2NO(g) +...
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