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For a rxn A+B+C -> D, the following results were obtained from 4 experiments. Write the rate law of the rxn, derive the rate law, derive the order of the reaction

Chemistry

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Answer

Rate=k[A][C]\text{Rate} = k[A][C]

k=30 M−1s−1k = 30 \text{ M}^{-1}\text{s}^{-1}

2nd order\text{2nd order}

Step-by-step solution

6 steps

The general rate law for the reaction A+B+C→DA + B + C \to D is given by:

Rate=k[A]x[B]y[C]z\text{Rate} = k[A]^x[B]^y[C]^z

where x,y,zx, y, z are the orders of reaction with respect to A, B, and C, respectively, and kk is the rate constant.

  1. Determine the order with respect to [B] (yy)

    Compare Experiment 1 and Experiment 2, where [A] and [C] are constant.

    Rate2Rate1=k[A]2x[B]2y[C]2zk[A]1x[B]1y[C]1z\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k[A]_2^x[B]_2^y[C]_2^z}{k[A]_1^x[B]_1^y[C]_1^z} 0.006 M s−10.006 M s−1=k(0.01 M)x(0.2 M)y(0.02 M)zk(0.01 M)x(0.01 M)y(0.02 M)z\frac{0.006 \text{ M s}^{-1}}{0.006 \text{ M s}^{-1}} = \frac{k(0.01 \text{ M})^x(0.2 \text{ M})^y(0.02 \text{ M})^z}{k(0.01 \text{ M})^x(0.01 \text{ M})^y(0.02 \text{ M})^z} 1=(0.20.01)y1 = \left(\frac{0.2}{0.01}\right)^y 1=(20)y1 = (20)^y

    This implies y=0y = 0.
    The reaction is zero order with respect to B.

  2. Determine the order with respect to [C] (zz)

    Compare Experiment 1 and Experiment 4, where [A] and [B] are constant.

    Rate1Rate4=k[A]1x[B]1y[C]1zk[A]4x[B]4y[C]4z\frac{\text{Rate}_1}{\text{Rate}_4} = \frac{k[A]_1^x[B]_1^y[C]_1^z}{k[A]_4^x[B]_4^y[C]_4^z} 0.006 M s−10.003 M s−1=k(0.01 M)x(0.01 M)y(0.02 M)zk(0.01 M)x(0.01 M)y(0.01 M)z\frac{0.006 \text{ M s}^{-1}}{0.003 \text{ M s}^{-1}} = \frac{k(0.01 \text{ M})^x(0.01 \text{ M})^y(0.02 \text{ M})^z}{k(0.01 \text{ M})^x(0.01 \text{ M})^y(0.01 \text{ M})^z} 2=(0.020.01)z2 = \left(\frac{0.02}{0.01}\right)^z 2=(2)z2 = (2)^z

    This implies z=1z = 1.
    The reaction is first order with respect to C.

  3. Determine the order with respect to [A] (xx)

    Compare Experiment 1 and Experiment 3. We know y=0y=0 and z=1z=1.

    Rate3Rate1=k[A]3x[B]3y[C]3zk[A]1x[B]1y[C]1z\frac{\text{Rate}_3}{\text{Rate}_1} = \frac{k[A]_3^x[B]_3^y[C]_3^z}{k[A]_1^x[B]_1^y[C]_1^z} 0.018 M s−10.006 M s−1=k(0.03 M)x(0.02 M)0(0.02 M)1k(0.01 M)x(0.01 M)0(0.02 M)1\frac{0.018 \text{ M s}^{-1}}{0.006 \text{ M s}^{-1}} = \frac{k(0.03 \text{ M})^x(0.02 \text{ M})^0(0.02 \text{ M})^1}{k(0.01 \text{ M})^x(0.01 \text{ M})^0(0.02 \text{ M})^1} 3=(0.030.01)x(0.020.02)13 = \left(\frac{0.03}{0.01}\right)^x \left(\frac{0.02}{0.02}\right)^1 3=(3)x(1)3 = (3)^x (1)

    This implies x=1x = 1.
    The reaction is first order with respect to A.

  4. Write the rate law (Question 1 and part of Question 2)

    Using the determined orders x=1,y=0,z=1x=1, y=0, z=1:

    Rate=k[A]1[B]0[C]1\text{Rate} = k[A]^1[B]^0[C]^1 Rate=k[A][C]\text{Rate} = k[A][C]

    The rate law is:

    Rate=k[A][C]\boxed{\text{Rate} = k[A][C]}
  5. Calculate the rate constant (kk) (Remaining part of Question 2)

    Using data from Experiment 1:
    Rate = 0.006 M s−10.006 \text{ M s}^{-1}
    [A]=0.01 M[A] = 0.01 \text{ M}
    [C]=0.02 M[C] = 0.02 \text{ M}

    0.006 M s−1=k(0.01 M)(0.02 M)0.006 \text{ M s}^{-1} = k(0.01 \text{ M})(0.02 \text{ M}) 0.006 M s−1=k(0.0002 M2)0.006 \text{ M s}^{-1} = k(0.0002 \text{ M}^2) k=0.006 M s−10.0002 M2k = \frac{0.006 \text{ M s}^{-1}}{0.0002 \text{ M}^2} k=30 M−1s−1k = 30 \text{ M}^{-1}\text{s}^{-1}

    The rate constant is:

    k=30 M−1s−1\boxed{k = 30 \text{ M}^{-1}\text{s}^{-1}}
  6. Derive the overall order of the reaction (Question 3)

    The overall order of the reaction is the sum of the individual orders:
    Overall order = x+y+z=1+0+1=2x + y + z = 1 + 0 + 1 = 2.
    The overall order of the reaction is:

    2nd order\boxed{\text{2nd order}}
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