Equilibrium constant, K_(c) " for the reaction" , N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) " at " 500 K " is " 0*061. At a particular time, the analysis shows that composition of the reaction mixture is 3*0" mol "L^(-1) N_(2), 2*0 " mol "L^(-1) H_(2)and5*0 "mol " L^(-1) NH_(3).Is the reaction at equilibrium ? If not , in which direction does the reaction tend to reach equilibrium ?

Equilibrium constant, K_(c) " for the reaction" , N_(2) (g) + 3 H_(2)

1.	Equilibrium constant, K_(c) " for the reaction" , N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) " at " 500 K " is " 0061. At a particular time, the analysis shows that composition of the reaction mixture is 30" mol "L^(-1) N_(2), 20 " mol "L^(-1) H_(2)and50 "mol " L^(-1) NH_(3).Is the reaction at equilibrium ? If not , in which direction does the reaction tend to reach equilibrium ?
Answer» Solution :`Q_(C)= ([NH_(3)]^(2))/([N_(2)][H_(2)]^(3) )=(05)^(2)/((30)(20)^(3))=00104` As ` Q_(c)!=K_(c) `, REACTION is not in equilibrium As ` Q_(c) ltK_(c)`, reaction willproceed in the forwarddirection.