8.) The following data was obtained for the reaction : Trials Initial [A1(M)0.25

Question

8.) The following data was obtained for the reaction : Trials Initial [A1(M)0.250-1-0.333 Initial[B1(M)0.333 | 0.250 Initial [C1 (M Initial Rate (M/s) 1.48x10 4.55x10 1 2 3 4 | 0.250 | 0.333 0.315 1.52x10 | 0.250 | 0.750 0.125 4.81x10s 0.145 0.500 a) Write the rate expression for this reaction in terms of product E. b) Write the rate law for this reaction (show work) c) Determine the overall order of the reaction. d) Calculate the rate constant. e) An additional trial is performed using a concentration of A that is two times that of B and half that of C. The observed rate is 3.99x106 M/s. Calculate the concentrations of A, B, and C used in this experiment.

Explanation / Answer

dE/dt= K[A]a [B]b [C]c, where a, b and c are orders of reaction with respect to A, B and C respectively.

E is thr product and K is the rate constant.dE/dt is the rate

From trial-1,1.48*10-5= K[0.25]a[0.333]b[ 0.145]c    (1)

Trial-2, 4.55*10-4 = K[0.333]a [0.25]b [0.5]c (2)

From trial-3, 1.52*10-4= K[0.25]a [0.333]b [0.315]c (3)

From trial-4, 4.81*10-5= K[0.25]a [0.75]b[0.125]c                      (4)

Eq.3/Eq.1 gives 1.52*10-4/ (1.48*10-5)= (0.315/0.145)c, 2.17c= 10.27, c=3

Eq.1 becomes with this value of C, 1.48*10-5= K[0.25]a [0.333]b*(0.145)3

0.0048546= K[0.25]a [0.333]b   (1A)

Eq.4 with the value of c becomes 4.81*10-5= K[0.25]a[0.75]b [0.125]3,

0.0246272= K[0.25]a [0.75]b   (4A)

Eq.4A/Ea.1A= 0.0246272/0.0048546= 5= (0.75/0.333)b, b= 2

Eq.1A now, 0.0048546= K[0.25]a (0.333)2= K[0.25]a= 0.043779   (1B)

Trial-2 now, 4.55*10-4 = K[0.333]a [0.25]2 [0.5]3 (2A)or K[0.333]a = 0.05824 (2A)

Eq.2A/Eq.1B gives (0.333/0.25)a= 0.05824/0.043779, a=1

Hence the Eq.2A becones K[0.333]=0.05824, K=0.174895/M5.s

So the rate law becomes dE/dt= 0.174895*[A]1[B]2[C]3,

Overall order is sum of orders of individual orders, overall order=1+2+3=6

Given concentration of A is two times that of B and half of that of C.

[B]=[A]/2 and [C]=2A

So the rate = 0.174895[A] [A/2] [2A]3= 3.99*10-6 M/s, [A]=5.7*10-6

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