For the reaction A + B + 2C right arrow 3 D at 25 degreeC the following initial

Question

For the reaction A + B + 2C right arrow 3 D at 25 degreeC the following initial rate data were obtained: Write the rate law expression for the reaction. What is a plausible rate-determining step in the mechanism of this reaction? What is the numerical value for k, the rate constant for this reaction? (Becareful of units) In Experiment 1, what would the rate of the reaction be when exactly 1/2 of the A present initially has been consumed? In experiment 2, what is the maximum concentration obtainable for D? (Assume that the volume remains constant.) In experiment 4, when the reaction has proceeded to completion, which reactant or reactants if any, and in what concentration, remain unconsumed?

Explanation / Answer

Answer – a) We are given the reaction and initial rate data

Reaction – A + B +2C ----> 3D

We assume the rate law for this reaction

Rate = k [A]x [B]y [C]z

The x, y and z are the order with respect to A,B and C

Rate1 = k [A-]1x [B]1y [C]1z

Rate2 = k [A-]2x [B]2y [C]2z

Rate3 = k [A-]3x [B]3y [C]3z

Rate4 = k [A-]4x [B]4y [C]4z

Now order with respect to B -

Rate3/ Rate1 = k [A]3x [B]3y [C]3z / k [A-]1x [B]1y [C]1z

4.0x10-4 / 4.0 x10-4 = (0.1)x /(0.1)x * (0.3)y /(0.10)y *(0.2)z /(0.2)z

1 = (3)y

y = 0

Now order with respect to A -

Rate2/ Rate1 = k [A-]2x [B]2y [C]2z / k [A-]1x [B]1y [C]1z

1.2 x10-3 / 4.0 x10-4 = (0.3)x /(0.1)x * (0.2)0 /(0.1)0 *(0.2)z /(0.2)z

3 = (3)x

x = 1

Now order with respect to C

Rate4 / Rate2 = k [A-]4x [B]4y [C]4z / k [A-]2x [B]2y [C]2z

3.6 x10-3 / 1.2*10-3 = (0.3)1 /(0.3)1 * (0.40)0 /(0.2)0 *(0.6)z /(0.2)z

3 = (3)z

z = 1

so rate law –

Rate = k [A] [C]

b) We determined the rate law and from that we confirm the rate of reaction is depending on the A and C and not on the B. The order with respect to A and C is 1, so in the rate determining step means slow step there is A and C with coefficient 1 and it I ass follow –

A + C ----> 2D   slow step ( Rate determining step)

B + C ----> D fast step

A + B +2C ----> 3D   Overall reaction

c) Calculation of rate constant –

We know rate law

Rate = k [A] [C]

4.0 x 10-4 M/min = k x (0.1)(0.20)

4.0 x 10-4 M/min = k x 0.02

So, k = 4.0 x 10-4 M/min / 0.02 M2

   k = 0.02 M-1 min-1

d) We are given, Exp-1, initial concentration, [A]o = 0.1 M , [C]o = 0.2 M

when gets half then [A] = 0.05 M , [C] = 0.1 M

we already calculated rate constant, so rate of reaction is

Rate = k [A] [C]

         = 0.02 M-1­­­.min-1 * (0.05 M) *(0.1 M)

          = 1.0*10-4 M/min

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