C + A D (slow) D F (fast) A. What is the actual rate law for this reaction? B. A
ID: 696498 • Letter: C
Question
C + A D (slow)
D F (fast)
A. What is the actual rate law for this reaction?
B. A reaction was run and found to have a half life that got shorter as the initial concentration of the reactant ([A]) was increased. When the reaction was run at two different temperatures, the following data sets were generated.
A. What is the actual rate law for this reaction?
B. A reaction was run and found to have a half life that got shorter as the initial concentration of the reactant ([A]) was increased. When the reaction was run at two different temperatures, the following data sets were generated.
At 550 K At 840 K [A] Time (sec.) [A] Time (sec.) 0.275 0.00 0.275 0.00 0.0673 255.0 0.00411 14.0 0.0340 585.0 0.00166 35.0 0.0198 1065.0 0.00124 47.0 0.00749 2950 0.000432 135.0 Use the data given to determine the value of k (the rate constant) at each temperature, and then determine the Energy of Activation for this reaction.Explanation / Answer
A. Let consider rate constant for the steps are k1, k-1, k2, and k3 respectively.
Then, d[F]/dt = k3[D] ; d[D]/dt = k2[C][A] - k3[D]
d[C}/dt =k1[A][B] - k-1[C] - k2[C][A]
Now applying steady state approximation on D, we will get, d[D]/dt = k2[C][A] - k3[D] =0
so, k3[D] = k2[C][A]
Again applying rate determinining state spproximation on C, we will get, k1/ k-1 = [C] / [A][B]
So, [C] = k1[A][B] / k-1
rate of formation of F, d[F]/dt = k3[D] = k2[C][A] = k2{ k1[A][B] / k-1} [A] = ( k1k2/k-1) [A]2[B]
Therefore, actual rate equation, d[F]/dt = ( k1k2/k-1) [A]2[B]
B. half life of a nth order reaction t1/2 = B/a(n-1) , where a is initial concentration.
So the the reaction can't be 1st order.
If we asseume the reaction as zero order reaction, then rate equation would be x = kt
Now using this equation if we put values of x and t for set at 550 K temp, then for each couple of data we will get rate constant value k~ 10-4 .
k = x/t
So when t =255 sec., x =(0.275 - 0.0673) = 10-4
again when t = 585 sec. then. x = (0.275-0.034) = 10-4
Hence it is a zero order reaction and value of rate constant at 550 K is 10-4 mol.-1L.s-1
Similarly for the set of 840 K , for each couple of data rate constant would be 10-3 mol.-1L.s-1
Now from Arrhenius equation, ln(k1/k2) = - Ea/R {1/T1 -1/T2}
In the above equation, putting k1 = 10-4 mol.-1L.s-1 , k2 = 10-3 mol.-1L.s-1, T1 = 550 K and T2 = 840 K, we get Ea = 30497.8 J/mol.= 30.49 KJ/mol.
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