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.

For the reaction N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g), DeltaH=?

Unit of equilibrium constant (K_c) for the reaction N_2(g)+3H_2(g)

The equilibrium of the reaction N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) will be shifted to the right when:

For the reaction, N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) , the units of K_(p) are …………

For the reaction, N_2[g] + 3H_2[g] hArr 2NH_3[g] , Delta H = …

For the reversible reaction N_(2) (g) +3H_(2) (g) hArr 2NH_(3) (g) + " Heat" The equilibrium shifts in forward direction

What is DeltaG^(ɵ) for the following reaction? 1/2 N_(2)(g)+3/2 H_(2)(g) hArr NH_(3)(g), K_(p)=4.42xx10^(4) at 25^(@)C

What is DeltaG^(ɵ) for the following reaction? 1/2 N_(2)(g)+3/2 H_(2)(g) hArr NH_(3)(g), K_(p)=4.42xx10^(4) at 25^(@)C

Write the relation between K_(p) " and " K_(c) for the reaction: N_(2)(g) +3H_(2) (g) hArr 2NH_(3)(g)

Equilibrium constant, K_(c) for the reaction, N_(2(g))+3H_(2(g))hArr2NH_(3(g)) , at 500 K is 0.061 litre^(2) "mole"^(-2) . At a particular time, the analysis shows that composition of the reaction mixture is 3.00 mol litre^(-1)N_(2) , 2.00 mol litre^(-1)H_(2) , and 0.500 mol litre^(-1)NH_(3) . Is the reaction at equilibrium? If not, in which direction does the reaction tend to proceed to reach equilibrium?