Question

For the multi-step reaction A+B → C+D the rate-limiting step is unimolecular, with A as the sole reactant. If (A) and (B) are both 0.195 M, then the rate of reaction is 0.0080 M/s. What is the rate of the reaction if (A) is doubled? rate: M/s HE Starting with the original concentrations, what is the rate of the reaction if B is doubled? M/s rate: Starting with the original concentrations, what is the rate of the reaction if(A) and B) are both increased by a factor of 3?

Answer 1

0 A + B +C+D Rate = K[A] .... 0 Now, Rate = 0.0080 M/S and [A]= 0.195 M = [A], Rale ak [4] , in K = Ratlo - 10.0080 ,-) . Lato 00195) 5. Now, [A] = 2x 0-195 M = 2 [A]. Rate = 0.0080 x 2x0.195 M/S = 0.0160 M/SCANS) 20:195 0 IN 2nd case, rate = 0.0080 Miss because the rate is . unimolecular with respect to the concentration of A only and the rate does not depends on the concentration of B7 . Rate = K[A Now, [4] = 3 [A]. = k {3[410]'{3 [B].] [B] = 5[]. = 3* K [4]. * 1 = 3x Rates = 3 x 0:00 80 M/S = 0.0240 M/S (ans)