The equilibrium constant for the formation of calcium carbonate from the ions in solution is 2.2*10⁸ according to the reaction: Ca²⁺(aq) + CO₃²^-(aq) CaCO₃(s)
What is the value of the equilibrium constant for the reverse of this reaction?

Question

The equilibrium constant for the formation of calcium carbonate from the ions in solution is 2.2*10⁸ according to the reaction: Ca²⁺(aq) + CO₃²^-(aq)

The equilibrium constant for the formation of calcium carbonate from the ions in solution is 2.2*10<sup>8</sup> according to the reaction: Ca<sup>2+</sup>(aq) + CO<sub>3</sub><sup>2</sup><sup>-</sup>(aq) CaCO<sub>3</sub>(s) What is the value of the equilibrium constant for the reverse of this reaction? A) the same, 2.2 * 10<sup>8</sup> B) -2.2 * 10<sup>8</sup> C) 2.2 * 10<sup>-</sup><sup>8</sup> D) 4.5 * 10<sup>-</sup><sup>9</sup> E) 4.5*10<sup>9</sup>

CaCO₃(s)
What is the value of the equilibrium constant for the reverse of this reaction?

Quizplus · Accepted Answer

The Answer of The equilibrium constant for the formation of calcium carbonate from the ions in solution is 2.2*10⁸ according to the reaction: Ca²⁺(aq) + CO₃²^-(aq) CaCO₃(s) What is the value of the equilibrium constant for the reverse of this reaction?

The Equilibrium Constant for the Formation of Calcium Carbonate from the Ions