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40% mixture of 0.2 mole of N2 and 0.6 mo...

40% mixture of 0.2 mole of `N_2` and 0.6 mole of `H_2` react to give `NH_3` according to the equation,
`N_(2(g))+ 3H_(2(g)) Leftrightarrow 2NH_(3(g))`
at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is

A

`4:5`

B

`5:4`

C

`7:10`

D

`8:5`

Text Solution

Verified by Experts

The correct Answer is:
A
`"Initial moles "underset(0.2)(N_(2(g))+underset(0.6)(3H_(2)(g)) Leftrightarrow underset(0)(2NH_(3(g))`
At equil. `0.2 -(0.2 xx 40)/(100)" "0.6-(0.6 xx 40)/(100)`
`" 0.2 -0.08=0.12 0.6-0.24 =0.36 0.16"`
Initial moles =0.2 +0.6=0.8
Final moles =0.12+0.36+0.16=0.64
pV =nRT
`V alpha n` (At constant temp and pressure)
`(0.64)/(0.8)=(640)/(800)=4/5`
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