A mixture of 1.57 mol of N_2 1.92 mol of H_2 and 8.13 mol of NH_3 is introduced into a 20 L reaction vessel at 500 K . At this temperature, the equilibrium constant K_c for the reactin. N_(2)(g)+3H_(2) (g)j hArr 2NH_(3)(g) is 1.7 xx 10^(-2) Is this reaction at equilibriuim ? if not , what is the direction of net rection ?

A mixture of 1.57 mol of N_2 1.92 mol of H_2 and 8.13 mol of NH_3 is i

1.	A mixture of 1.57 mol of N_2 1.92 mol of H_2 and 8.13 mol of NH_3 is introduced into a 20 L reaction vessel at 500 K . At this temperature, the equilibrium constant K_c for the reactin. N_(2)(g)+3H_(2) (g)j hArr 2NH_(3)(g) is 1.7 xx 10^(-2) Is this reaction at equilibriuim ? if not , what is the direction of net rection ?
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :The reaction is `N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g)`<br/> Conecentration quotient `(Q_c)= ([NH_3]^)/([N_2][H_2]^3)` <br/> `=((8.13/20 " mol "L^(-1))^1)/((1.57/20 "mol " L^(-1))xx(1.92/20 " mol "L^(-1))^3)=2.38 xx <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^3`<br/> The equilibrium contant `(K_c)` for the reaction `=1.7 xx 10^(-2)`<br/> As `Q_(<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>) <a href="https://interviewquestions.tuteehub.com/tag/ne-1112263" style="font-weight:bold;" target="_blank" title="Click to know more about NE">NE</a> K_(C)` , this means that the reaction is not in a state of equilibrium.</body></html>