A) B) C) The reaction A + B rightarrow C + D rate = k[A][B]2 has an initial rate

Question

A)

B)

C)

The reaction A + B rightarrow C + D rate = k[A][B]2 has an initial rate of 0.0940 M/s. What will the initial rate be if [A] is halved and [B] is triped? What will the initial rate be if [A] is triped and [B] is halved? Consider the following mechanism. step; 2A B equilibrium step: B + C rightarrow D slow overall: 2A + C rightarrow D Determine the rate law for the overall reaction (where the overall rate constant is represented as K). rate = Using the given data, determine the rate constant of this reaction. A + 2B rightarrow C + D

Explanation / Answer

A) Assume that the initial concentrations of each reactant are both 0.5 M. Then, from the rate law and the initial

rate, you can calculate k as:

rate = k(0.5)(0.5)^2 = 0.0940

k = 0.752 M^-2s^-1

Now, if you have half [A], it would be 0.25M and triple [B] to 1.5M. The rate then would be:

rate = 0.752 (0.25)(1.5)^2 = 0.423 M/s

Just to make sure this works, let's assume that the initial concentrations of both were 0.25 M. In that instance, k

would be,

0.094 = k (0.25)(0.25)^2

k = 6.016

And if you halve A and triple B, you would have

rate = 6.016 (0.125)(0.75)^2 = 0.423 M/s

So, under the first set of concentrations, if you triple [A] and halve [B],

rate = 0.752 (1.5)(0.25)2 = 0.0705 M/s

B) Lets us rewrite the equations by putting rate constants for each,

2A <====> B assuming k1 is the rate constant for forward reaction and k-1 is rate cosntant for reverse reaction.

B + C ------> D assume k2 is rate constant for this reaction.

Since this is the slow step, the rate of the reaction is dictated by this slow step,

Thus the rate law can be written as,

Rate = k2[B][C]

Now we cannot have intermediate concentration [B] in the Rate law, so

Finding [B] from the first equation.

Rate of forward reaction = Rate of reverse reaction

That is, k1[A]2 = k-1[B]

Rearranging for [B] = k1/k-1[A]2

Substituting this value in the first rate law equation we get,

Rate = k2[(k1/k-1[A]2].[C]

        = k[A]2[C] is the rate law for the overall reaction

Wherein, k = k2.k1/k-1

C) Looking at the table we see that,

When you double the concentration of B (Trial 2) from Initial Trial run 1, the rate does not change. Therefore, the reaction is zero order with respect to [B], and [B] will not appear in the rate law expression.

Since doubling [A] as in Trial run 3 increases the rate of the reaction 4-fold from the intial Trial run 1, the reaction is second order with respect to A. So, the rate law is:

rate = k[A]^2

Feeding the intial rate and concentration of A into the equation we get,

0.0211 = k(0.250)^2

Solving for k we get,

k = 0.338 M^-1 s^-1

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