A)With initial concentrations of A and B of 0.04 mol/L, the rate of formation of

Question

A)With initial concentrations of A and B of 0.04 mol/L, the rate of formation of AB is measured at 0.005 mmol/L. When the initial concentration of A is raised to 0.07 mol/L with that of B is the same as before, the rate of formation of AB increased to 0.015 mmol/L, What is the order of reaction for A?......................... B)The reaction is found to be third order overall; what is order in B?........................ C) If initial concentration of B is changed to 0.08 mol/L while that of A is reduced to 0.02 mol/L what will be the new rate of formation of AB?................... D) For a certain set of initial concentrations of A and B, it takes 10 sec to produce 0.1 mmol of AB at 20 C at 150 sec at 0 C. What is the ratio of the rate at 20 C to that of 0 C?....................... I think the answer to A is 2nd order and B is 1st order.. C) 0.0025mm/L x S but I'm not sure.. could you please tell me if im right and if not how to do it and Im completely confused on D!!! Thank you!!!

Explanation / Answer

A) rate 2/rate 1 = 0.015/0.005 = [[A2]/[A1] ]^n , 3 = ( 0.07/0.04)^n = 3 , n = 2 , hence order with respect to A = 2 , (B) total order = 3 = oder A + order B , hence order of B = 3-order A = 3-2 = 1, (C) rate = k(0.04)^2(0.04) = 0.005 , k = 78.125 , so rate when A = 0.08 , B= 0.02 is rate = 78.125(0.08)^2(0.02) = 0.01 , (D) rate 2/rate 1 = k2/k1 , we have ln(k2/k1) = ( E/R)(1/T1- 1/T2) , where T1 = C = 273 K , T2 = 20C = 293 K , we need to have Ea to get ratio of k2/k1 , then rate 2/rate 1 = k2/k1

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