2. (20 points) The initial rates for the reaction of A and B under various initi

Question

2. (20 points) The initial rates for the reaction of A and B under various initial conditions are listed below. Determine the reaction order for reactant A. (A plotting grid is provided on the next pageTyou wouldlike toplotsomething. In AM B M initial rate M/s) 0.00200 0.00200 O - -c, 0.00132- 0.0810 vs t 0.185 0.002000448 400 5,91 0.0025o 55y0.00389 0.0 o 0.4062.19 J , 0.00200 0.07 0.00200 0.00254 20.195.. 0.00200- 0.00309 0.129 200 ~ 0.00510 0.231 thial rate-Slope s rport ona to order order [BJ is' Arst order

Explanation / Answer

In order to calculate the rate law expression for a A+B reaction, we need to apply Initial Rates Method.

Note that the generic formula goes as follows:

r = k [A]^a [B]^b

Note that if we got at least 3 sets of point, in which we have A and B constant, then we could use:

r1 / r2 = (k1 [A]1^a [B]1^b) / (k2 [A]2^a [B]2^b)

If we assume K1 and K2 are constant, then K1= K2 cancel each other

r1 / r2 = ([A]1^a [B]1^b) / ( [A]2^a [B]2^b)

Then, order according to [A] and [B]

r1 / r2 = ([A]1/[A2])^a * ([B]1/[B]2)^b

If we get two points in which A1 = A2, then we could get B, and vise versa for A...

From the data shown in YOUR table

Choose point:

1 and 2

(0.081)/(0.185) = (0.00132/0.0025)^a

a = ln(0.4378) / ln(0.528) = 1.29

choose point 3 and 4 to confirm

(0.406)/(0.448) = (0.00389/0.00417)^a

a = ln(0.90625) / ln(0.93285) = 1.41

Verify with point 2 and 4

(0.185)/(0.448) = (0.0025/0.00417)^a

a = ln(0.4129) / ln(0.5995) = 1.72

now, choose last 2 points

(0.231/0.384) = (0.00309/0.00510)^b

b = ln((0.231/0.384)) / ln((0.00309/0.00510))

b = 1

then

Rate = k*[A]^a * [B]^b

Rate = k*[A]^1.5 [B]

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