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1 mol of N(2) is mixed with 3 mol of H(2...

`1` mol of `N_(2)` is mixed with `3` mol of `H_(2)` in a litre container. If `50%` of `N_(2)` is converted into ammonia by the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`, then the total number of moles of gas at the equilibrium are

A

`1.5`

B

`4.5`

C

`3.0`

D

`6.0`

Text Solution

Verified by Experts

The correct Answer is:
C
`alpha=50%=50/100=0.5`
`{:(,N_(2),+,3H_(2),hArr,2NH_(3)),("Initial",1,,3,,0),("Final",1-alpha,,3-3alpha,,2alpha),("Total moles"=1-alpha+3-3alpha+2alpha,,,,,):}`
`=4-2alpha`
`=4-2xx0.5=3`
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