Determine `K_(c )` for the reaction
`(1)/(2)N_(2)(g)+(1)/(2)O_(2)(g)+(1)/(2)Br_(2)(g)hArrNOBr(g)`
from the following data at `298 K`.
The equilibrium constants for the following reaction
`2NO(g)hArrN_(2)(g)+O_(2)(g)`
and `NO(g)+(1)/(2)Br_(2)(g)hArrNOBr(g)`
are `2.4xx10^(30)` and `1.4`, respectively.

Explanation

the net reaction is
`(1)/(2)N_(2)(g)+(1)/(2)O_(2)(g)+(1)/(2)Br_(2)`
`K_(C _(("net")))=([NOBr])/([N_(2)]^(1//2)[O_(2)]^(1//2)[Br_(2)]^(1//2))`
Considering the given equations:
`2NO(g)hArrN_(2)(g)+O_(2)(g),`
Equilibrium constant `=2.4xx10^(30)`
`N_(2)(g)+O_(2)(g)hArr2NHO(g),`
Equilibrium constant`=1/(2.4xx10^(30))`
`(1)/(2)N_(2)(g)+(1)/(2)O_(2)(g)hArrNO(g),`
Equlibrium constant `=(1/(2.4xx10^(30)))^(1//2)=0.6455xx10^(-15)`
`([NO])/([N_(2)]^(1//2)[O_(2)]^(1//2))=K_(c )^(')=0.6455xx10^(-15) ...(i)`
`NO(g)+(1)/(2)Br_(2)(g)hArrNOBr(g)`
`([NOBr])/([NO][Br_(2)]^(1//2))=K_(c )^('')=1.4 ...(ii)`
Multiply both equaton
`([NO])/([N_(2)]^(1//2)[O_(2)]^(1//2))xx([NOBr])/([NO][Br_(2)]^(1//2))`
`K_(C )^(')xxK_(C )^('')=0.6455xx10^(-15)xx1.4`
or `([NO])/([N_(2)]^(1//2)[O_(2)]^(1//2)[Br_(2)]^(1//2))=K_(C _(("net")))=0.9037xx10^(-15)`
`=9.037xx10^(-16)`