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2 "mole" N(2) and 3 "mole" H(2) gas are ...

`2 "mole" N_(2)` and `3 "mole" H_(2)` gas are allowed to react in a `20 L` flask at `400 K` and after complete conversion of `H_(2)` into `NH_(3)`. `10 L H_(2)O` was added and temperature reduced to `300 K`. Pressure of the gas after reaction is :
`N_(2)+3H_(2)to2NH_(3)`

A

`3Rxx(300)/(20)`

B

`3Rxx(300)/(10)`

C

`Rxx(300)/(20)`

D

`Rxx(300)/(10)`

Text Solution

Verified by Experts

The correct Answer is:
D
`N_(2)` left `=1 mol`
`NH_(3)` formed`=2 mol` but dissolved in `H_(2)O`, hence pressure is due to `N_(2)` only Hence, volume of the flask is `10 L` since 10 L `H_(2)O` is added.
`thereforeP=(n)/(V)RT=(300R)/(10)`
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