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.

Explanation

To find the concentration of \( N_2 \) at equilibrium for the reaction:

\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

given the equilibrium constant \( K_c = 6.0 \times 10^{-2} \) at 715 K, and the concentrations of \( H_2 \) and \( NH_3 \) at equilibrium, follow these steps:

...