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.

Explanation

To solve the problem, we need to use the equilibrium constant expression for the given reaction:

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

**Given:**
- Equilibrium constant \( K_c = 6.0 \times 10^{-2} \) at 715 K
- Concentration of \( H_2 = 0.25 \, \text{mol L}^{-1} \)
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