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Answer:

rate = k [ CH 3 CHO ] 2 with k = 6.73 × 10 −6 L/mol/s

Determining rate laws from initial rates

Using the initial rates method and the experimental data, determine the rate law and the value of the rate constant for this reaction:

2NO( g ) + Cl 2 ( g ) 2NOCl( g )
Trial [NO] (mol/L) [Cl 2 ] (mol/L) Δ [ NO ] Δ t ( mol L −1 s −1 )
1 0.10 0.10 0.00300
2 0.10 0.15 0.00450
3 0.15 0.10 0.00675

Solution

The rate law for this reaction will have the form:

rate = k [ NO ] m [ Cl 2 ] n

As in [link] , we can approach this problem in a stepwise fashion, determining the values of m and n from the experimental data and then using these values to determine the value of k . In this example, however, we will use a different approach to determine the values of m and n :

  1. Determine the value of m from the data in which [NO] varies and [Cl 2 ] is constant. We can write the ratios with the subscripts x and y to indicate data from two different trials:

    rate x rate y = k [ NO ] x m [ Cl 2 ] x n k [ NO ] y m [ Cl 2 ] y n

    Using the third trial and the first trial, in which [Cl 2 ] does not vary, gives:

    rate 3 rate 1 = 0.00675 0.00300 = k ( 0.15 ) m ( 0.10 ) n k (0.10) m ( 0.10 ) n

    After canceling equivalent terms in the numerator and denominator, we are left with:

    0.00675 0.00300 = ( 0.15 ) m ( 0.10 ) m

    which simplifies to:

    2.25 = ( 1.5 ) m

    We can use natural logs to determine the value of the exponent m :

    ln ( 2.25 ) = m ln ( 1.5 ) ln ( 2.25 ) ln ( 1.5 ) = m 2 = m

    We can confirm the result easily, since:

    1.5 2 = 2.25

  2. Determine the value of n from data in which [Cl 2 ] varies and [NO]is constant.

    rate 2 rate 1 = 0.00450 0.00300 = k ( 0.10 ) m ( 0.15 ) n k ( 0.10 ) m ( 0.10 ) n

    Cancelation gives:

    0.0045 0.0030 = ( 0.15 ) n ( 0.10 ) n

    which simplifies to:

    1.5 = ( 1.5 ) n

    Thus n must be 1, and the form of the rate law is:

    Rate = k [ NO ] m [ Cl 2 ] n = k [ NO ] 2 [ Cl 2 ]

  3. Determine the numerical value of the rate constant k with appropriate units. The units for the rate of a reaction are mol/L/s. The units for k are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate. In this example, the concentration units are mol 3 /L 3 . The units for k should be mol −2 L 2 /s so that the rate is in terms of mol/L/s.

    To determine the value of k once the rate law expression has been solved, simply plug in values from the first experimental trial and solve for k :

    0.00300 mol L 1 s −1 = k ( 0.10 mol L −1 ) 2 ( 0.10 mol L −1 ) 1 k = 3.0 mol −2 L 2 s −1

Check your learning

Use the provided initial rate data to derive the rate law for the reaction whose equation is:

OCl ( a q ) + I ( a q ) OI ( a q ) + Cl ( a q )
Trial [OCl ] (mol/L) [I ] (mol/L) Initial Rate (mol/L/s)
1 0.0040 0.0020 0.00184
2 0.0020 0.0040 0.00092
3 0.0020 0.0020 0.00046

Determine the rate law expression and the value of the rate constant k with appropriate units for this reaction.

Answer:

rate 2 rate 3 = 0.00092 0.00046 = k ( 0.0020 ) x ( 0.0040 ) y k ( 0.0020 ) x ( 0.0020 ) y
2.00 = 2.00 y
y = 1
rate 1 rate 2 = 0.00184 0.00092 = k ( 0.0040 ) x ( 0.0020 ) y k ( 0.0020 ) x ( 0.0040 ) y
2.00 = 2 x 2 y 2.00 = 2 x 2 1 4.00 = 2 x x = 2
Substituting the concentration data from trial 1 and solving for k yields:
rate = k [ OCl ] 2 [ I ] 1 0.00184 = k (0.0040) 2 (0.0020) 1 k = 5.75 × 10 4 mol 2 L 2 s 1

Reaction order and rate constant units

In some of our examples, the reaction orders in the rate law happen to be the same as the coefficients in the chemical equation for the reaction. This is merely a coincidence and very often not the case.

Practice Key Terms 5

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Source:  OpenStax, Ut austin - principles of chemistry. OpenStax CNX. Mar 31, 2016 Download for free at http://legacy.cnx.org/content/col11830/1.13
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