The equilibrium constant for the reaction:
`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)` at `725K`
is `6.0xx10^(-2)`. At equilibrium `[H_(2)]=0.25 "mol" L^(-1)` and `[NO_(3)]=0.06 "mol" L^(-1)`
Calculate the equilirbium concentration of `N_(2)`.

The equilibrium constant for the reaction : N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)" at 715 K is "6.0xx10^(-2) . If, in a particular reaction, there are 0.25" mol L"^(-1) of H_(2) and 0.06" mol L"^(-1) of NH_(3) present, calculate the concentration of N_(2) at equilibrium.

The equilibrium constant for the reaction : N_2(g)+3H_2(g) hArr 2NH_3(g) at 715 K is 6.0xx10^(-2) If in a particular reaction , there are 0.25 mol L^(-1) of H_2 and 0.06 mol L^(-1) of NH_3 present at equilibrium , calculate the concentration of N_2 at equilibrium.

The equilibrium constant for the reaction, H_(2)(g) + I_(2)(g) hArr 2HI(g) is 64 at a certain temperature. The equilibrium concentrations of H_(2) and HI are 2 mol//L and 16 mol//L respectively. What is the equilibrium concentration ( "in "mol//L ) of I_(2) ?

The equilibrium constant K_(P) for the reaction N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) is 8.19xx10^(2) at 298K and 4.6xx10^(-1) at 498 K. Calculate DeltaH^(@) fror the reaction.