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To which direction the equillibrium shif...

To which direction the equillibrium shifts when `N_(2)` is added at equilibrium condition ? `N_(2) + 3H_(2) hArr 2NH_(3)`

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The correct Answer is:
B

Because change in Gubbs free energy is given by `Delta G` (reaction ) = `Delta G` (products) - `Delta G` (reactants). Also, catalyst only speeds up the reaction without affecting the equilibrium conditions.
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The following concentration were obtained for the formation of NH_(3)(g) from N_(2)(g) and H_(2)(g) at equilibrium and at 500 K: [N_(2)] = 1 xx 10^(-2)M, [H_(2)] = 2 xx 10^(-2)M and [NH_(3)] = 2xx 10^(-2)M . The equilibrium constant, K_(c) , for the reaction N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g) at 500 K is

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The following equilibria and their equilibrium constants are given N_(2)+3H_(2) iff 2NH_(3)(K_(1)) N_(2)+O_(2) iff 2NO(K_(2)) H_(2)+1/2O_(2) iff H_(2)O(K_(3)) Therefore, the equilibrium constant of the reaction 2NH_(3)+5/2O_(2) iff 2NO+3H_(2)O in terms of K_(1), K_(2) and K_(3) is :

BRILLIANT PUBLICATION-CHEMICAL AND IONIC EQUILIBRIUM-Level -II
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