Rate law equation: For the reaction A+B+CD+E, the initial reaction rate was meas

Question

Rate law equation:

For the reaction A+B+CD+E, the initial reaction rate was measured for various initial concentrations of reactants. The following data were collected:

Trial [A] (M) [B] (M) [C] (M) Initial rate (M/s)

1 0.20 0.20 0.20 6.0*10^-5

2 0.20 0.20 0.60 1.8*10^-4

3 0.40 0.20 0.20 2.4*10^-4

4 0.40 0.40 0.20 2.4*10^-4

The rate of a chemical reaction depends on the concentrations of the reactants. For the general reaction between Aand B,

aA+bBcC+dD

The dependence of the reaction rate on the concentration of each reactant is given by the equation called the rate law:

rate=k[A]m[B]n

where k is a proportionality constant called the rate constant. The exponent m determines the reaction order with respect to A, and n determines the reaction order with respect to B. The overall reaction order equals the sum of the exponents (m+n).

Part D: What is the value of the rate constant k for this reaction? Express your answer to two significant figures and include the appropriate units. Indicate the multiplication of units explicitly either with a multiplication dot or a dash.

Explanation / Answer

Let the rate law be :-

rate of reaction = k*[A]m*[B]n

Now, 6*10-5 = k*(0.2)m*(0.2)n ......(1)

2.4*10-4 = k*(0.4)m*(0.4)n .....(2)

2.4*10-4 = k*(0.4)m*(0.2)n .....(3)

Now, (2)/(3) ; gives

1 = (2)n

or, n = 0

Also, (3)/(1) gives

4 = 2m

or, m = 2

Thus, putting the values of m & n in (1) we get

6*10-5 = k*(0.2)2

or,rte constant, k = 1.5*10-3 M-1s-1

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