Question

Consider the reaction. A(aq)↽−−⇀2B(aq)Kc=1.83×10^−6 at 500 K If a 2.90 M sample of A is heated to 500 K, what is the concentration of B at equilibrium?

Answer 1

To determine the concentration of B at equilibrium for the reaction A(aq) ⇌⇋ 2B(aq), we can use an ICE table and the provided equilibrium constant (Kc = 1.83 × 10⁻⁶) at 500 K. The ICE table accounts for the change in concentrations of A and B from initial to equilibrium states, which can be used to solve for the concentration of B.

Step-by-step explanation:

This college-level chemistry question involves equilibrium calculations which are a key concept in chemical kinetics and thermodynamics. The reaction described is A(aq) ⇌⇋ 2B(aq) with an equilibrium constant (Kc) of 1.83 × 10⁻⁶ at 500 K. If we start with 2.90 M of A, and the reaction reaches equilibrium at 500 K, we can determine the concentration of B at equilibrium using an ICE (Initial, Change, Equilibrium) table.

Let's define the concentrations at equilibrium:
Initial: [A] = 2.90 M, [B] = 0 M (since the reaction hasn't started yet)
Change: [A] will decrease by x, [B] will increase by 2x (since 2 moles of B are produced for every mole of A that reacts)
Equilibrium: [A] = 2.90 - x, [B] = 2x

Using the equilibrium constant expression for the reaction:
Kc = [B]² / [A] = (2x)² / (2.90 - x) = 1.83 × 10⁻⁶
This equation can be solved for x to find the concentration of B at equilibrium.

Answer 2

To calculate the equilibrium concentration of B, an ICE table is used along with the equilibrium expression obtained from the given Kc value. By solving for x, which represents the change in concentration, the equilibrium concentration of B is determined. Due to the low Kc value, the equilibrium concentration of B will be low.

Step-by-step explanation:

The question asks to calculate the concentration of B at equilibrium for the reaction A(aq) ⇌⇋ 2B(aq) with a given equilibrium constant Kc and an initial concentration of A. To find the equilibrium concentration of B, we use an ICE table (Initial, Change, Equilibrium) to express the concentrations of A and B at equilibrium.

Let x be the change in concentration of A as it becomes 2B, so the change for B would be 2x. At equilibrium, the concentration of A will be 2.90 M - x, and the concentration of B will be 2x. Using the given Kc = 1.83×10⁻⁶ at 500 K, we have the following equilibrium expression:

Kc = [B]^2 / [A]

We plug in the expressions in terms of x to get:

1.83×10⁻⁶ = (2x)^2 / (2.90 - x)

This equation is then solved for x to find the concentration of B at equilibrium. Finally, since the reaction strongly favors the reactants at a high equilibrium constant, we expect the change x to be very small relative to the initial concentration of A, which means that B at equilibrium will be very low compared to the initial concentration of A.