1.

The reaction N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) is in equlibrium. Now the reaction mixture is compressed to half the volume

Answer»

More of ammonia will be formed
Ammonia will dissociate back into `N_(2)and H_(2)`
There will be no effect on equlibrium
Equlibrium constant of the reaction will change

Solution :`K_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3)).` What volume is halved, CONCENTRATIONS are DOUBLED. To keep `K_(c)` CONSTATN, increase of `[NH_(3)]` should be more.


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