For the reaction
`N_(2) + 3 H_(2) = 2 NH_(3)`,

For the reaction N_(2) + 3 H_(2) hArr 2 NH_(3) equilibrium constant is K_(1) and that of for the reaction NH_(3) hArr (1)/(2) N_(2) + (3)/(2) H_(2) is K_(2) . (ii) Then calculate the relation between K_(1) and K_(2) .

For the reaction N_(2) + 3 H_(2) to 2 NH_(3) The rate of change of concentration for hydrogen is -0.3 xx 10^(-4) Ms^(-1) The rate of change of concentration of ammonia is :

For the reaction N_(2) + 3 H_(2) to 2 NH_(3) The rate of change of concentration for hydrogen is 0.3 xx 10^(-4) Ms^(-1) The rate of change of concentration of ammonia is :

For the reaction N_(2) + 3 H_(2) to 2 NH_(3) The rate of change of concentration for hydrogen is 0.3 xx 10^(-4) Ms^(-1) The rate of change of concentration of ammonia is :

For the reaction, N_(2) + 3H_(2) hArr 2NH_(3) in a vessel, equal moles of N_(2) and H_(2) are mixed to attain equilibrium.

For the reaction N_(2) + 3H_(2) to 2NH_(3) if (Delta[NH_(3)])/(Deltat) = 2 xx 10^(-4) mol L^(-1)s^(-1) , the value of (-Delta[H_(2)])/(Deltat) would be

For the reaction , N_(2)+3H_(2) to 2NH_(3), "if" (d[NH_(3)])/(dt)=2xx10^(-4)"mol"L^(-1)s^(-1), the value of (-d[H_(2)])/(dt) would be