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16 August, 04:28

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]

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  1. 16 August, 08:03
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    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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