For the reaction:2 A (g) B (g) + C (g)at 900 ° C, Kc is 1.40 x 10 -3. If 0.780 mole of A (g) and 0.244 mole each of B (g) and C (g) are mixed in a1.00 liter container at 900 ° C, what are the concentrations of A, B, and C at equilibrium?

To find the concentrations of A, B, and C at equilibrium, we can use the given equilibrium constant (Kc) and the initial moles of each gas.

Step 1: Write the balanced chemical equation for the reaction: 2 A (g) -> B (g) + C (g)

Step 2: Calculate the initial concentrations of A, B, and C: Concentration of A = moles of A / volume of container = 0.780 mol / 1.00 L = 0.780 M Concentration of B = moles of B / volume of container = 0.244 mol / 1.00 L = 0.244 M Concentration of C = moles of C / volume of container = 0.244 mol / 1.00 L = 0.244 M

Step 3: Use the stoichiometry of the reaction to determine the change in concentrations at equilibrium: Since the stoichiometric coefficient of A is 2, the change in concentration of A will be -2x. Since the stoichiometric coefficients of B and C are both 1, the change in concentrations of B and C will be +x.

Step 4: Substitute the initial concentrations and the changes in concentrations into the equilibrium expression: Kc = [B] * [C] / [A]^2 1.40 x 10^-3 = (0.244 + x) * (0.244 + x) / (0.780 - 2x)^2

Step 5: Solve the quadratic equation for x: 1.40 x 10^-3 = (0.244 + x)^2 / (0.780 - 2x)^2

Solving this equation will give the value of x, which represents the change in concentration of A, B, and C at equilibrium.

Step 6: Calculate the equilibrium concentrations of A, B, and C: Concentration of A at equilibrium = 0.780 - 2x Concentration of B at equilibrium = 0.244 + x Concentration of C at equilibrium = 0.244 + x

By substituting the value of x into these equations, we can find the equilibrium concentrations of A, B, and C.