A 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)iff2NH_(3)(g)` at consta

Question

A 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)iff2NH_(3)(g)` at consta

A 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)iff2NH_(3)(g)` at constant temperature & pressure. 40 % mole of nitrogen react at equilibrium state. Then the ratio of the final volume to the initial volume of gases are
A. `4 : 5`
B. `5 : 4`
C. `7 : 10`
D. `8 : 5`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - A
`N_(2)(g)+3H_(2)(g)iff2NH_(3)(g)`
mole `" 0.2 0.6 "`
Left mole `" 0.12 0.36 0.16"`
`therefore " "(V_(f))/(V_(i n))=(0.64)/(0.80) " "(because "T and P constant")`
`=0.8" or "(4 : 5)`

4 views Answered 2023-02-05 15:29