occurred chemical reaction scheme: A + 2B = B. The starting materials and the reaction product - gases.At some point a balance, that is, the rate of the forward reaction (A + 2B = B) caught up with the speed of the back (B = A + 2B).It is known that the equilibrium concentration of substance A is equal to 0.12 mol / liter, the element B - 0.24 mol / liter and the substance B - 0.432 mol / liter.It is required to determine the initial concentration of A and B.
Explore scheme chemical interaction.It follows that one mole of the product
(item B) formed from one mole of a substance A and two moles of substance B. If in one liter of the reaction volume was formed 0.432 mol The item (under the terms of the problem), then, respectively, at the same time it consumed 0.432 mole of a substanceA and 0.864 mol of element B.
you know the equilibrium concentrations of the starting materials: [A] = 0,12 mol / liter, [B] = 0.24 mol / liter.Adding to these quantities those that were consumed during the reaction, you get the value of the initial concentrations: [A] 0 = 0.12 + 0.432 = 0.552 mol / liter;[B] 0 = 0.24 + 0.864 = 1.104 mol / liter.
You can also determine the initial concentration of substances with the help of the equilibrium constant (Kp) - The ratio of product equilibrium concentrations of reaction products to the product of the equilibrium concentrations of the starting materials.The equilibrium constant is calculated by the formula: Cr = [C] n [D] m / ([A] 0x [B] 0y), where [C] and [D] - equilibrium concentrations of reaction products, and D;n, m - coefficients thereof.Accordingly, [A] 0, [B] 0 - equilibrium concentrations of the elements entering into the reaction;x, y - their coefficients.
Knowing the exact scheme of the ongoing reaction, the equilibrium concentration at least one product and starting material, as well as the value of the equilibrium constant, we can write this problem in terms of a system of two equations with two unknowns.