Question

> Activities and Due Dates > HW 11 core: 350/1800 Resources Hint 2 of 18 > sing the data in the table, calculate the rate constant of is reaction. Trial [A] (M) [B] (M) Rate (M/s) 0.390 0.370 0.0179 0.103 A+B- C+D 0.888 0.390 0.624 0.370 0.0286 R 1 Y U TOP F G H J K L : V B N M

Answer 1

rin si Given that Exp. [A][M] 0.390 2. 0.390 3. 0.624 [B][M] 0.370 0.888 0.370 Initial rate (M/s) 0.0179 0.103 0.0286 The rate law is Rate = k [A] * [B] Y where, x = order of reaction with respect to [A] y= order of reaction with respect to [B] From exp.1 and 2 [A] = constant, So rate = k [B] Y From. 1: 0.0179 = k (0.370)y ........eq.(1) From.2: 0.103 = k (0.888) ........eq.(2) Divide eq.2 with eq.1 0.103 k(0.888) 0.0179 k(0.370)" 35.75=(2.4) (2.4)' = (2.4) y=2

From exp.1 and 3 [B] = constant, So rate = k [A] * From. 1: 0.0179 = k (0.390)* From.2: 0.0286 = k (0.624)* Divide eq.3 with eq.4 ........eq.(3) ........eq.(4) 0.0286 k(0.624)* 0.0179 k(0.390) 31.60 =(1.6)* >(1.6)' = (1.6)* > x=1 The rate law is Rate = k [A] * [B] Y = k [A]” [B]2 = k [A][B]? k=- TA rate [A][B] From Exp.1, 0.0179 M.s-1 2 = 0.335262 M2 5-1 (0.390M )(0.370M) k = 0.335 M².5-1 k=-