Question

Consider the reaction: A(aq)⇌2B(aq) At 500 K, K=8.31×10 −6 . If a 3.90 M sample of A is heated to 500 K, what is the concentration of B at equilibrium? Option 1: 1.95 M Option 2: 0.13 M Option 3: 0.26 M Option 4: 3.77 M

Answer 1

In the reaction A(aq)⇌2B(aq), and given K = 8.31*10^-6 at 500 K, if we start with a 3.90 M sample of A, the concentration of B at equilibrium would be 0.13 M.

Step-by-step explanation:

This problem is about calculating the concentration of B at equilibrium. Given the balanced chemical equation, A(aq)⇌2B(aq), we can define the change in concentrations from the initial to equilibrium conditions using the ICE table (Initial, Change, Equilibrium). Initially, we have 3.90 M of A and 0 M of B. At equilibrium, assuming x mol/L of A dissociates, we end up with (3.90-x) M of A and 2x M of B. The equilibrium constant K is given by [B]^2/A.

Plugging in these quantities into K expression: K = [B]^2/[A] = (2x)^2 / (3.90-x) = 8.31*10^-6. This is a quadratic equation of form ax^2 + bx + c = 0. Solving this equation for x yields x ~0.065.

However, we're interested in the concentration of B, which is 2x. Therefore, the concentration of B at equilibrium is 2*0.065, which equals 0.13 M. So, Option 2 is the correct answer.

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