Question

`DeltaG^(ɵ)=77.77 kJ mol^(-1)` at `1000 K` for the reaction `1//2 N_(2)(g)+1//2O_(2)(g) hArr NO(g)`. What is the partial pressure of `NO` under equilibrium at `1000 K` for air at `1 atm` pressure containing `80% N_(2)` and `20% O_(2)` volume.

Answer 1

`DeltaG^(ɵ)=-2.303RT log K_(c )`
`:. 77.77xx10^(3)=-2.303xx8.314xx1000 log K_(c )`
`:. K_(c )=8.67xx10^(-5)`
`{:(,1//2 N_(2),+,1//2 O_(2),hArr,NO),("Initial pressure",0.8,,0.2,,0),("Final pressure",(0.8-x//2),,(0.2-x//2),,x):}`
`K_(c )=([NO])/([N_(2)]^(1//2)[O_(2)]^(1//2))`
`:. 8.67xx10^(-5)=(x)/((0.8-x//2)^(1//2)(0.2-x)^(1//2))`
`x=3.47xx10^(-5) "atm"`