Given the following data table, use the method of initial rates to determinet th

ID: 801314 • Letter: G

Question

Given the following data table, use the method of initial rates to determinet the rate law for the following reaction: A + B = C + D + E

run: ro(M/s): [A]o: [B]o: [C]o: [D]o: [E]o:

1 4.62 1.00 2.00 0.500 3.00 0.25

2 4.62 2.00 2.00 1.00 3.00 0.25

3 2.31 1.00 2.00 1.00 3.00 0.500

4 18.5 2.00 4.00 1.00 3.00 0.500

5 13.1 1.00 4.00 0.500 6.00 0.250

6 13.1 1.00 4.00 0.500 6.00 0.500

Let the rate of reaction be r = k [A]^a [B]^b [C]^c [D]^d [E]^e

From 5 and 6,

when concentration of E is doubled, the rate remains unchanged. Hence, the rate is independent of E

Hence, e= 0

From 1 and 2, when A is doubled and C is also doubled, the rate of reaction is unchanged.

Hence, a+c = 0 or a = -c

From 1 and 3, when C is doubled, rate is halved {NOTE: We have already proved that reaction is independent of e }

Thus, 0.5 = 2^c

c = -1

thus, a = 1

from 1 and 5, when B is doubled, D is also doubled

13.1 / 4.62 = [2^b][2^d]

b + d= 1.50

consider, 1 and 4, only B is doubled

thus , 18.5 / 4.62 = 2^b

b = 2

d = -0.50

Thus, the rate law becomes:

r = K [A] [B]^2 / [C] [D]^0.5

Substituting the values:

4.62 = K [1.00] [2.00]^2 / [0.5] [3.00]^0.5

4.62 = 4.62 K

Thus, k = 1

Thus, the final rate law =>

r = [A] [B]^2 / [C] [D]^0.5

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