Question

Write equilibrium constant \((K)\) expressions for the following reactions: (a) \(\mathrm{I}_{2}(s)+2 \mathrm{Cl}^{-}(g) \rightleftharpoons \mathrm{Cl}_{2}(g)+2 \mathrm{I}^{-}(a q)\) (b) \(\mathrm{CH}_{3} \mathrm{NH}_{2}(a q)+\mathrm{H}^{+}(a q) \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}(a q)\) (c) \(\mathrm{Au}^{2+}(a q)+4 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Au}(\mathrm{CN})_{4}^{2-}(a q)\)

Short answer

#Question# Write the equilibrium constant expressions for the following reactions: (a) $$\mathrm{I}_{2}(s)+2 \mathrm{Cl}^{-}(g) \rightleftharpoons \mathrm{Cl}_{2}(g)+2 \mathrm{I}^{-}(a q)$$ (b) $$\mathrm{CH}_{3} \mathrm{NH}_{2}(a q)+\mathrm{H}^{+}(a q) \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}(a q)$$ (c) $$\mathrm{Au}^{2+}(a q)+4 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Au}(\mathrm{CN})_{4}^{2-}(a q)$$ #Answer# (a) $$K_{a} =\frac{[\mathrm{I}^{-}]^2}{[\mathrm{Cl}^{-}]^2}$$ (b) $$K_{b} = \frac{[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}]}{[\mathrm{CH}_{3} \mathrm{NH}_{2}][\mathrm{H}^{+}]}$$ (c) $$K_{c} = \frac{[\mathrm{Au}(\mathrm{CN})_{4}^{2-}]}{[\mathrm{Au}^{2+}][\mathrm{CN}^{-}]^4}$$

Step-by-step solution

(Reaction 1: Equilibrium constant expression)

Given reaction (a): $$\mathrm{I}_{2}(s)+2 \mathrm{Cl}^{-}(g) \rightleftharpoons \mathrm{Cl}_{2}(g)+2 \mathrm{I}^{-}(a q)$$ The equilibrium expression can be written as: $$K_{a} = \frac{[\mathrm{Cl}_{2}][\mathrm{I}^{-}]^2}{[\mathrm{I}_{2}][\mathrm{Cl}^{-}]^2}$$ However, since \(\mathrm{I}_{2}\) is in solid state, its concentration remains constant and does not affect the position of the equilibrium. Thus, we can simplify the expression as: $$K_{a} =\frac{[\mathrm{I}^{-}]^2}{[\mathrm{Cl}^{-}]^2}$$

(Reaction 2: Equilibrium constant expression)

Given reaction (b): $$\mathrm{CH}_{3} \mathrm{NH}_{2}(a q)+\mathrm{H}^{+}(a q) \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}(a q)$$ The equilibrium expression can be written as: $$K_{b} = \frac{[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}]}{[\mathrm{CH}_{3} \mathrm{NH}_{2}][\mathrm{H}^{+}]}$$

(Reaction 3: Equilibrium constant expression)

Given reaction (c): $$\mathrm{Au}^{2+}(a q)+4 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Au}(\mathrm{CN})_{4}^{2-}(a q)$$ The equilibrium expression can be written as: $$K_{c} = \frac{[\mathrm{Au}(\mathrm{CN})_{4}^{2-}]}{[\mathrm{Au}^{2+}][\mathrm{CN}^{-}]^4}$$