mical Kinetics Favorites Question 6 Not yet answered Marked out of 1.00 The init

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mical Kinetics Favorites Question 6 Not yet answered Marked out of 1.00 The initial rates for a reaction A and B were studied using the following initial concentrations in mo/L Expt 1 (A]-0.200, (BJ-0.200, rate-4.29x10 M/s Ext 2. [A]=0.307, [B1-0.200, rate-8.52x10; MS Expt 3. [A]-0.397, [B]-0.409, rate 3.56x10 M/s What is the value of the rate constant? .2 P Flag question Select one oa 0.429 M s o b. 0.536 MS 2 -1 2 -1 C. 0.429 M -. d. 0.536 M 1 -1 S Question 7 Not yet answered Marked out of The initial rates for a reaction A and B were studied using the following initial concentrations in mol/L -3 Ext 1 . [A]=0.100, [B]=0.400, rate= 1.6 1x10: Mis Expt 2. [A]=0.152, [B]-o400, rate245x1Q "MS Expt 3. [AJ-0.152, [B-1.57, rate-9.62x10 Ms What is the value of the rate constant? 1.00 P Flag question Select one 2 -1 2 -1 c. 0.0402 M s a.0.0442 MS b. 0.0402 M s 1 -1 d, 0.0442 M.' s. Question 8 Not yet answered Marked out of 1.00 P Flag question The initial rates for a reaction A and B were studied using the following Initial concentrations in mol/L. -3 Expt 1 . [A1-0.200, [B-0.200, rate-429x100 MS Expt 2. [A]-0.397, (B]-0.200, rate-1.69x10 M/s Expt 3. [AJ-0.397, (Bl-0.409, rate -3.46x10 Mis What is the order of reaction with respect to A? Select one: b. 0 c. 2

Explanation / Answer

6.

Do (1) / (2)

4.29 * 10-3 / ( 8.52 * 10-3) = (0.200 / 0.397)m

0.504 = (0.504)m

So, m = 1

Therefore, order with respect to A = 1

Now do (2) / (3)

8.52 * 10-3 / (3.56 * 10-2) = (0.200 / 0.409)n

0.239 = (0.489)n

(0.489)2 = (0.489)n

n = 2

Therefore, order with respect ot B = 2

Hence, Rate law expresion can be written as,

Rate = k[A][B]2

From (1)

4.29 * 10-3 = k (0.200)(0.200)2

k = 0.536 M-2s-1

so (b) is the answer as it is third order reaction in over all.

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