The reaction `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)` is in equlibrium. Now the reaction mixture is compressed to half the volumeA. More of ammonia will be formedB. Ammonia will dissociate back into `N_(2)and H_(2)`C. There will be no effect on equlibriumD. Equlibrium constant of the reaction will change

The reaction `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)` is in equlibrium. Now

1.	The reaction `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)` is in equlibrium. Now the reaction mixture is compressed to half the volumeA. More of ammonia will be formedB. Ammonia will dissociate back into `N_(2)and H_(2)`C. There will be no effect on equlibriumD. Equlibrium constant of the reaction will change
Answer» Correct Answer - A `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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The reaction `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)` is in equlibrium. Now the reaction mixture is compressed to half the volumeA. More of ammonia will be formedB. Ammonia will dissociate back into `N_(2)and H_(2)`C. There will be no effect on equlibriumD. Equlibrium constant of the reaction will change

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