The concentration of E_2 in the reaction below is doubled and the observed effec

Question

The concentration of E_2 in the reaction below is doubled and the observed effect on the rate is noted. E_2 + CHE_3 rightarrow CE_4 + HE Match the observed changed in rate with the reaction order for reagent E_2. the reaction rate increases by a factor of four. the reaction rate increases by a factor of six. the following set of data was obtained by the method of initial rates for the reaction: A_2B_2 (g) rightarrow A_2 (g) + B_2 (g) run a: [A] = 0.228 M. rate = 0.165 M/s run b: [A] = 0.465 M. rate = 0.709 M/s Determine the order of the reaction and then calculate and report the average value of the rate constant.

Explanation / Answer

SOLUTION:

Q1.

(A) Let the order w.r.t E2 is n, the rate increass by four times if concentration is doubled

Rate (II)/ Rate (I) = (2E2 /E2)n

4 / 1 = 2n

Or n = 2 hence rate w.r.t E2 is 2

(B) Rate becomes six fold

Rate (II)/ Rate (I) = (2E2 /E2)n

6 / 1 = 2n

log6 = nlog

0.778 = n X 0.301; n = 2.58

Q2.

Rate = - (A2B2) / dt = (A2) / dt = 0.165 ======(1)

Or dt = (A2) / 0.165 = dt = 0.228 / 0.165 = 1.38s

Rate = - (A2B2) / dt = 0.165

dt = 1.38s

- (A2B2) / 1.38s = 0.165

- (A2B2) = 1.38s X 0.165 = 0.227M

Similarly from (b) dt =  - 0.465 / 0.709 = 0.655 s

- (A2B2) = 0.655 X 0.709 = 0.464M

Rate (II) / Rate (I) = [(A2)/(A2)]n

0.709 / 0.165(I) = [0.464/0.227]n

4.29 = (2.044)n

n = 2, hence order is 2.

Rate (I) = K[A2B2]2

0.165 = K (0.227)2

K = 0.165 / 0.05 = 3.3

Similarly Rate (II) = K[A2B2]2

0.709 = K (0.464)2

K = 3.297

Average K = 3.297 + 3.3 / 2 = 3.298

Dear candidate please the other question separately as per guidelines one question is to be solved one time

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