10 mole of `NH_(3)` is heated at 15 atm from `27^(@C` to `347^(@)C` assuming volume constant. The pressure at equilibrium is found to be 50 atm. The equilibrium constant for dissociation of `NH_(3)` :
`2NH_(3) hArr N_(2) + 3H_(2), Delta H = 91.94 kJ` can be written as `K_(p) = (p_(N_(2)) xx (p_(H_(2)))^(2))/((P_(NH_(3))^(2)))("atm")^(2)`
The degree of dissociation of `NH_(3)` is :

Options

A. `61.3%`
B. `20%`
C. `48%`
D. `50%`

Answer

A

Explanation

`{:(2NH_(3),hArr,N_(2)+,3H_(2)),(100,,00,"00 at 300 K"),((10-2x),,x x,"3x at 620 K"):}`
Pressure increase due to increases in temperature as well as due to increase in mole.
Initiall `P prop T` Given `T_(1) = 300 K, P_(1) = 15` atm, `T_(2) = 620 K`
`P_(2) = (620)/(300) xx 15 = 31` atm, Now , `NH_(3)` is dissociated to attain 50 atm at 620 K. Thus, `P prop n` or `10 prop 31`
`100 + 2 x prop 50 rArr 2x = 6.13 therefore alpha = (2x)/(10) xx 100 = (6.13 xx 100)/(10) = 61.3%`