In a container equilibrium `N_(2)O_(4) (g)hArr2NO_(2) (g)` is attained at `25^(@)C`. The total equilibrium pressure in container is 380 torr. If equilibrium constant of above equilibrium is `0.667` atm, then degree of dissociation of `N_(2)O_(4)` at this temperature will be:

Options

A. `1//3`
B. `1//2`
C. `2//3`
D. `1//4`

Answer

B

Explanation

`" "N_2O_4(g) harr 2NO_2(g)`
`{:("initially, t=0", " "1, " "0),(t=t_(eq), " "1-alpha, " "2alpha):}`
`K_p=(4alpha^2)/(1-alpha^2)" P=380 torr"=(380)/(760)= .5 atm`
`0.667=(4alpha^2)/(1-alpha^2)xx1/2=0.667=(4alpha^2)/(1-alpha^2). 1/2, (4alpha^2)/(1-alpha^2)=0.667xx2`
`alpha=1/2`