At 25°C,the equilibrium constant K_c for the reaction
2A(aq)
B(aq) + C(aq)
Is 65.If 2.50 mol of A is added to enough water to prepare 1.00 L of solution,what will the equilibrium concentration of A be?

Question

At 25°C,the equilibrium constant K_c for the reaction
2A(aq)

At 25°C,the equilibrium constant K<sub>c</sub> for the reaction 2A(aq) B(aq) + C(aq) Is 65.If 2.50 mol of A is added to enough water to prepare 1.00 L of solution,what will the equilibrium concentration of A be? A) 0.038 M B) 0.14 M C) 0.28 M D) 1.18 M E) 2.4 M

B(aq) + C(aq)
Is 65.If 2.50 mol of A is added to enough water to prepare 1.00 L of solution,what will the equilibrium concentration of A be?

Quizplus · Accepted Answer

First, we need to set up the ICE (initial, change, equilibrium) table: | | A | B | C | |---------------|----------------|----------------|----------------| | Initial (M) | [A]₀ | 0 | 0 | | Change (M) | -2x | +x | +x | | Equilibrium (M)| [A]₀-2x | x | x | Note that we use the stoichiometric coefficients to determine the changes in concentration. Since 2 moles of A react to form 1 mole of B and 1 mole of C, we subtract 2x from [A]₀ and add x to both [B] and [C]. Now, we can use the equilibrium constant expression to solve for x: K_c = [B][C]/[A]^2 65 = x^2/([A]₀-2x)^2 Assuming x << [A]₀, we can approximate [A]₀-2x as [A]₀: 65 = x^2/[A]₀^2 x = [A]₀sqrt(65) We know that we added 2.50 moles of A to prepare the solution, and the volume of the solution is 1.00 L, so the initial concentration of A is: [A]₀ = 2.50 mol/1.00 L = 2.50 M Substituting into the equation for x, we get: x = 2.50 M * sqrt(65) = 11.8 M This value of x is clearly unreasonable, indicating that our assumption that x << [A]₀ is invalid. We need to use an iterative method to solve for x. Starting with an initial guess of x = 0.1 M: 65 = x^2/([A]₀-2x)^2 x = 0.1 M * ([A]₀-2x)/sqrt(65) x = 0.139 M This value of x is consistent with our assumption that x << [A]₀, so we can use it to calculate the equilibrium concentration of A: [A] = [A]₀-2x = 2.50 M - 2(0.139 M) = 2.22 M However, the answer choices are all given in units of M, so we need to convert this to M: [A] = 2.22 mol/1.00 L = 0.22 M Therefore, the correct answer is (B) 0.14 M.

At 25°C,the Equilibrium Constant Kc for the Reaction

At 25°C,the Equilibrium Constant K_c for the Reaction