2 mol `N_2` and 3 mol `H_2` are allowed to react in a 20 L flask at 400 K and after the complete conversion of `H_(2)` to `NH_(3), 10 L H_(2)O` was added and the temperature is reduced to 300 K. The pressure of gas after the reaction is `N_(2) + 3H_(2) to 2NH_(3)` (assume that all the `NH_(3)` formed gets dissolved in water)

Options

A. `3R xx 300/20`
B. `3R xx 300/10`
C. `R xx 300/20`
D. `R xx 300/10`

Answer

D

Explanation

`N_(2)` left = 1 mol
`NH_(3)` formed = 2 mol, but is dissolved in `H_2O`
Hence, pressure is only due to `N_2`. As volume of flask is 10 L, we add `10 L H_(2)O`.
`therefore P = n/V RT = (300 R)/10`