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The equilibrium constant for the reactio...

The equilibrium constant for the reaction:
`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)` at `725K`
is `6.0xx10^(-2)`. At equilibrium `[H_(2)]=0.25 "mol" L^(-1)` and `[NO_(3)]=0.06 "mol" L^(-1)`
Calculate the equilirbium concentration of `N_(2)`.

Text Solution

Verified by Experts

`underset(a)(N_(2))(g)+underset(0.25)(3H_(2))(g)hArr underset(0.06 "equilibrium concentration")(2NH_(3)(g))` ltbr. `K_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))`
`6.0xx10^(-2)=([0.06]^(2))/(axx(0.25)^(3))`
`a=([0.06]^(2))/(6.0xx10^(-2)xx(0.25)^(3))=3.84 "mol"L^(-1)`
`:.` The conc. of `N_(2)` at equilibrium `=3.84 "mol" L^(-1)`
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