Consider the following data showing the initial rate of a reaction at several di

Question

Consider the following data showing the initial rate of a reaction at several different concentrations of A. [A](M) 0.15 0.30 0.60
Initial Rate (M/s) 0.026 0.207 1.655
What is the order of the reaction? Express as an integer.
What is the rate law?
Estimate the value of the rate constant, k. Consider the following data showing the initial rate of a reaction at several different concentrations of A. [A](M) 0.15 0.30 0.60
Initial Rate (M/s) 0.026 0.207 1.655
What is the order of the reaction? Express as an integer.
What is the rate law?
Estimate the value of the rate constant, k. [A](M) 0.15 0.30 0.60
Initial Rate (M/s) 0.026 0.207 1.655
What is the order of the reaction? Express as an integer.
What is the rate law?
Estimate the value of the rate constant, k.

Explanation / Answer

The initial rate of the reaction means that the influence of the products on the reaction rate is negligible.
Hence, the rate is :
r = k[A]n where "n" being the order of the reaction.

log r = logK[A]n

log r = logK +´nlog[A]

The order can be determined by plotting log(r) vs log([A]), the slope being the order. In this case:

y = a + bx Where a = logK b = slope = n, x = log A; and y = log r

So if R2 is near 1, then the reaction is of the order indicated order:

log A: -0.8239; -0.5229; -0.2218

log r: -1.5850; -0.6840; 0.2188

a = log K = 0.88317

R2 = 0.999999996

b = n = 2.9961 or 3.

So the reaction order is 3.

Now that we know this the rate law would be:

r = k[A]3

Finally the value of K (two ways to do it):

a) 0.88317 = logK ---> k = 100.99317 = 7.6413

b) k = r/[A]3 = 0.026 / 0.153 = 7.704

Two really close values.

Hope this helps

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