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The following data were collected for the net reaction A + B2 + 2 C → D. Initial Initial Initial Initial [A] [B2] [C] rate M M M M/s 1 0.01 0.01 0.10 1.20 × 103 2 0.02 0.01 0.10 4.80 × 103 3 0.03 0.01 0.20 2.16 × 104 4 0.04 0.02 0.10 3.84 × 104 Which of the following is the rate law for this reaction? (Note that the units for the rate constant are omitted in the following answers.) 1. Rate = (1.2 × 1010) [A] [B2] 2 2. Rate = (1.2 × 109 ) [A] [B2] 2 3. Rate = (1.2 × 1011) [A]2 [B2] 2 4. Rate = (1.2 × 1010) [A]2 [B2] [C] 5. Rate = (1.2 × 106 ) [B2] [C] 6. Rate = (1.2 × 1012) [A]2 [B2] [C]

Respuesta :

Answer:

4. Rate = (1.2 × 1010) [A]2 [B2] [C]

Explanation:

A + B2 + 2 C → D

Initial Initial Initial Initial [A] [B2] [C] rate M M M M/s

1 0.01 0.01 0.10 1.20 × 103

2 0.02 0.01 0.10 4.80 × 103

3 0.03 0.01 0.20 2.16 × 104

4 0.04 0.02 0.10 3.84 × 104

Comparing experiment 1 and 2, when the concentration of a Is doubled, the rate increases by a factor of 4. This means the rate is second order with respect to A.

This reduces the options to;

3. Rate = (1.2 × 1011) [A]2 [B2] 2

4. Rate = (1.2 × 1010) [A]2 [B2] [C]

6. Rate = (1.2 × 1012) [A]2 [B2] [C]

From experiment 1, 2 , 3 and 4. Doubling the concentration of B increases the rate of equation by a factor of 2. This means the reaction is first order with respect to B.

This reduces the options to;

4. Rate = (1.2 × 1010) [A]2 [B2] [C]

6. Rate = (1.2 × 1012) [A]2 [B2] [C]

From the above options, we can conclude that the reaction is first order with respect to C.

Rate law is given as;

Rate = k  [A]2 [B2] [C]

From experiment 1;

1.20 × 103  = k (0.01)^2 (0.01) (0.1)

k = 1.20 × 103 / (1e-7)

k = 1.20 × 10^10

Hence, answer:

4. Rate = (1.2 × 1010) [A]2 [B2] [C]

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