If \(\mathrm{K}_{\mathrm{C}}\) for the formation of ammonia is \(2 \mathrm{moles}^{-2} \ell^{2}\), \(\mathrm{K}_{\mathrm{c}}\) for decomposition of ammonia is ___________.

The balanced equation for the formation of ammonia is: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

The decomposition of ammonia is the reverse reaction of its formation, so the balanced equation for the decomposition of ammonia is: 2NH₃(g) ⇌ N₂(g) + 3H₂(g) Since the decomposition of ammonia is the reverse reaction of its formation, the equilibrium constants are inversely related. If we denote the equilibrium constant for the formation of ammonia as \(\mathrm{K_{C_{formation}}}\) and the equilibrium constant for the decomposition of ammonia as \(\mathrm{K_{C_{decomposition}}}\), then: \(\mathrm{K_{C_{decomposition}}} = \frac{1}{\mathrm{K_{C_{formation}}}}\) Given, \(\mathrm{K_{C_{formation}}} = 2 \, \mathrm{moles}^{-2}\ell^{2}\). Substituting the value in the equation above, we get: \(\mathrm{K_{C_{decomposition}}} = \frac{1}{2 \, \mathrm{moles}^{-2}\ell^{2}}\)

To find \(\mathrm{K_{C_{decomposition}}}\), simplify the expression: \(\mathrm{K_{C_{decomposition}}} = \frac{1}{2 \, \mathrm{moles}^{-2}\ell^{2}}\) \(\mathrm{K_{C_{decomposition}}} = 0.5 \, \mathrm{moles}^{2}\ell^{-2}\) So, the equilibrium constant for the decomposition of ammonia is \(0.5 \, \mathrm{moles}^{2}\ell^{-2}\).