In a vessel with a volume of 8.0 L, 0.25 mol of substance A2 and 0.45 mol of substance B2 were mixed

In a vessel with a volume of 8.0 L, 0.25 mol of substance A2 and 0.45 mol of substance B2 were mixed at a certain temperature. After 25 minutes, an equilibrium was established in the system Ar (g) + B2 (g) = 2AB (g). The average consumption rate of B2 during this period was 1.0-10 ^ -3 mol / (L * min). Calculate the equilibrium constant and mole fractions of all substances in the equilibrium mixture. How many times is the speed of the direct reaction at equilibrium less than the initial speed (consider the direct reaction elementary)?

A2 (g) + B2 (g) = 2AB (g).
Let’s calculate the concentration of the starting substances
C [A2] start = 0.25 / 8 = 0.03125 mol / l
C [B2] start = 0.45 / 8 = 0.05625 mol / l
At a consumption rate of 10 ^ -3 mol / (L * min) in 25 minutes, the concentration of B2 will change by 10 ^ -3 * 25 = 0.025 mol / L. The concentration of A2 substance will change by 0.025 mol / l:
C [A2] = 0.03125 – 0.025 = 0.00625 mol / L.
C [B2] = 0.05625 – 0.025 = 0.03125 mol / L.
And the concentration of the AV substance will be:
C [AB] = 0.025 * 2 = 0.05 mol / L.
Then Kp = [AB] ^ 2 / ([A2] * [B2]) = 0.0025 / 1.95 * 10 ^ -4 = 12.82.
Ctot = 0.00625 + 0.03125 + 0.05 = 0.0875 mol / L.
χ [A2] = 0.0625 / 0.0875 = 0.0714.
χ [B2] = 0.03125 / 0.0875 = 0.3571.
χ [AB] = 0.05 / 0.0875 = 0.5714.
Forward reaction speed:
Vstraight = k * C [A2] * C [B2].
Then the ratio of the reaction rate in the equilibrium mixture to the initial mixture rate:
Vstart / Vequal = (k * C [A2] start * C [B2] start) / (k * C [A2] equals * C [B2] equals) =
= (0.25 * 0.45) / (0.0625 * 0.03125) = 576
Answer: it will decrease by 576 times.