A reaction A B has the following time dependence for the concentration of [A] vs
ID: 1023441 • Letter: A
Question
A reaction A B has the following time dependence for the concentration of [A] vs time. For t=(0 s, 5 s, 10 s, 15 s, 25 s) the concentration of [A]=(30.00 M, 20.01 M, 13.35 M, 8.90 M, 3.96 M). The initial concentration of [A] is the value at t=0 s.
(A)Calculate the values of the rate constant k assuming that the reaction is first order.What is the value of k at 5 s? s-1What is the value of k at 10 s? s-1What is the value of k at 15 s? s-1What is the value of k at 25 s? s-1
(B)Calculate the value of k assuming that the reaction is second order.What is the value of k at 5 s? M-1s-1What is the value of k at 10 s? M-1s-1What is the value of k at 15 s? M-1s-1What is the value of k at 25 s? M-1s-1
(C)Use your results from parts A and B above to decide what the order of the reaction is for your data. (Enter 1 or 2)
(D)What is the concentration of [A] at time t=50 s? M
Explanation / Answer
A reaction A B has the following time dependence for the concentration of [A] vs time. For t=(0 s, 5 s, 10 s, 15 s, 25 s) the concentration of [A]=(30.00 M, 20.01 M, 13.35 M, 8.90 M, 3.96 M). The initial concentration of [A] is the value at t=0 s.
(A)Calculate the values of the rate constant k assuming that the reaction is first order.What is the value of k at 5 s? s-1What is the value of k at 10 s? s-1What is the value of k at 15 s? s-1What is the value of k at 25 s? s-1
Answer:
(i) at t= 5 s
K = 2.303 /t ( log [A0 / At] )
K = 2.303 log [30 / 20.10] / 5 = 0.080
(ii) at t = 10 s
K = 2.303 /t ( log [A0 / At] )
K = 2.303 log [30 / 13.35] / 10 = 0.080
(iii) at t= 15 seconds
K = 2.303 /t ( log [A0 / At] )
K = 2.303 log [30 / 8.90] / 15 = 0.081
(iv) at t= 25 seconds
K = 2.303 /t ( log [A0 / At] )
K = 2.303 log [30 / 3.96] / 25 = 0.081
(B)Calculate the value of k assuming that the reaction is second order.What is the value of k at 5 s? M-1s-1What is the value of k at 10 s? M-1s-1What is the value of k at 15 s? M-1s-1What is the value of k at 25 s? M-1s-1
For second order reaction
1/[At] - 1/[A0] = Kt
(i) at t = 5 seconds
1/[At] - 1/[A0] = Kt
K = (1/[At] - 1/[A0] ) / time
K = (1/ 20.01 - 1/30 ) / 5
K = (0.049- 0.033 ) / 5 = 0.0032
(ii) at t= 10 seconds
1/[At] - 1/[A0] = Kt
K = (1/[At] - 1/[A0] ) / time
K = (1/ 13.35 - 1/30 ) / 10
K = (0.0749- 0.033 ) / 10 = 0.0042
(iii) at t= 15 seconds
1/[At] - 1/[A0] = Kt
K = (1/[At] - 1/[A0] ) / time
K = (1/ 8.90 - 1/30 ) / 15
K = (0.112- 0.033 ) / 15 = 0.0058
(iv) at t= 25 seconds
1/[At] - 1/[A0] = Kt
K = (1/[At] - 1/[A0] ) / time
K = (1/ 3.96 - 1/30 ) / 25
K = (0.252- 0.033 ) / 25 = 0.0088
C) from above data a constant value of rate constant is obtained for first order rate law so the reaction is first order reaction
D) at t= 50 seconds
K = 2.303 /t ( log [A0 / At] )
K = 2.303 log [30 / At] / 50 = 0.080
0.080 X 50 / 2.303 = log [30 / At]
1.736 = log [30 / At]
Taking antilog
54.45 = 30 / At
At = 30 / 54.45 = 0.55 M
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