Question

A reaction obeys the rate law: rate = k [A]2[B]. At some set of concentrations, the rate is found to be 8.2 x 10-2 M/sec. What will be the rate of reaction if the concentration of A is halved and the concentration of B is doubled? A. 4.1 x 10-2 M/sec B. 8.2 x 10-2 M/sec C. 5.0 x 10-1 M/sec D. 1.6 x 10-1 M/sec

Answer 1

Rate law : Rate =k[A]²[B] | Rate = .8.2x10² m/s = X [A]²[B] =R (het) Mrow concentration of A is halved so it becomes [A22] concentration of Bis doubled it becomes = [23] so R' = New rate = K[A] [28] = 2. KCAJ?[B] = 1 CAD CO] ICA]²[B]= R = 8.2x10²M/ So New rate, R' = Neu b = 8.2 x 10 ml 8.2 x 10 2 R' = 4.1 x 10² MA Oblien ca) is correct.