`K_1and K_2` are equilibrium constant for reactions (i) and (ii)
`N_2(g) +O_2(g)hArr2NO(g)` ....(i)
`NO(g)hArr(1)/(2)N_2(g)+(1)/(2)O_2(g)`...(ii)
Then,

Options

A. `1,4.1 xx 10^(-2) dm^(-3)atm^(-1)mol, 606 dm^(6)atm^(2)mol^(-2)`
B. `1,24.62" dm atm "mol^(-1), 606 dm^(6)atm^(2)mol^(-2)`
C. `24.62 dm^(3)atm mol^(-1), 606 dm^(6)atm^(2)mol^(-2)`
D. `1,24.62 dm^(3)atm mol^(-1), 1.65 xx 10^(-3) dm^(6)atm^(2)mol^(2)`

Answer

D

Explanation

`K_(p) = K_(c)(RT)^(Deltang)`
`K_(p)/K_(c) = (RT)^(Deltang)`
For reaction `N_(2)(g) + O_(2)(g) rarr 2NO(g)`
`Deltang = 0 hence Kp//Kc = 1`
For reaction `N_(2)O_(4)(g) rarr 2NO_(2)(g)`
`Deltang = 1` hence `Kp//Kc = (24.62)^(1) dm^(3) atm mol^(-1)`
For reaction `N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g)`
`Deltang = -2 hence Kp//Kc = (24.62)^(-2) = 1.65 xx 10^(-3) dm^(-6)atm^(-2)mol^(2)`