To solve this problem, we need to use the formula for the equilibrium constant (K) for the reaction. The equilibrium constant expression for the reaction is:

K = [C]^2 / ([A][B])

Given in the problem, we know that the initial moles of A and B are 1.0 mol and 3.5 mol respectively. Since the volume of the container is 1 L, these are also the initial concentrations in Molarity (M).

At equilibrium, the concentration of A is 0.8 M. This means that 0.2 M of A has reacted.

From the balanced chemical equation, we can see that for every mole of A that reacts, one mole of B also reacts and 2 moles of C are produced. Therefore, at equilibrium, the concentration of B is 3.5 M - 0.2 M = 3.3 M and the concentration of C is 2 * 0.2 M = 0.4 M.

Substituting these equilibrium concentrations into the equilibrium constant expression gives:

K = [C]^2 / ([A][B]) = (0.4 M)^2 / ((0.8 M)(3.3 M)) = 0.16 M^2 / 2.64 M^2 = 0.06

So, the value of K to 2 decimal places is 0.06.

Question

To solve this problem, we need to use the formula for the equilibrium constant (K) for the reaction. The equilibrium constant expression for the reaction is:

K = [C]^2 / ([A][B])

Given in the problem, we know that the initial moles of A and B are 1.0 mol and 3.5 mol respectively. Since the volume of the container is 1 L, these are also the initial concentrations in Molarity (M).

At equilibrium, the concentration of A is 0.8 M. This means that 0.2 M of A has reacted.

From the balanced chemical equation, we can see that for every mole of A that reacts, one mole of B also reacts and 2 moles of C are produced. Therefore, at equilibrium, the concentration of B is 3.5 M - 0.2 M = 3.3 M and the concentration of C is 2 * 0.2 M = 0.4 M.

Substituting these equilibrium concentrations into the equilibrium constant expression gives:

K = [C]^2 / ([A][B]) = (0.4 M)^2 / ((0.8 M)(3.3 M)) = 0.16 M^2 / 2.64 M^2 = 0.06

So, the value of K to 2 decimal places is 0.06.

Knowee AI · Accepted Answer

To solve this problem, we need to use the formula for the equilibrium constant (K) for the reaction. The equilibrium constant expression for the reaction is:

K = [C]^2 / ([A][B])

Given in the problem, we know that the initial moles of A and B are 1.0 mol and 3.5 mol respectively. Since the volume of the container is 1 L, these are also the initial concentrations in Molarity (M).

At equilibrium, the concentration of A is 0.8 M. This means that 0.2 M of A has reacted.

From the balanced chemical equation, we can see that for every mole of A that reacts, one mole of B also reacts and 2 moles of C are produced. Therefore, at equilibrium, the concentration of B is 3.5 M - 0.2 M = 3.3 M and the concentration of C is 2 * 0.2 M = 0.4 M.

Substituting these equilibrium concentrations into the equilibrium constant expression gives:

K = [C]^2 / ([A][B]) = (0.4 M)^2 / ((0.8 M)(3.3 M)) = 0.16 M^2 / 2.64 M^2 = 0.06

So, the value of K to 2 decimal places is 0.06.

Samples of A (1.0 mol) and B (3.5 mol) are placed in a 1 L container and the following reaction takes place: A(g) + B(g) ⇄ 2C(g) At equilibrium, the concentration of A is 0.8 M. What is the value of K to 2 d.p?

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