Constants| Periodic Table Part A The integrated rate laws for zero-, first-, and
ID: 1037870 • Letter: C
Question
Constants| Periodic Table Part A The integrated rate laws for zero-, first-, and second- order reaction may be arranged such that they resemble the equation for a straight line,y ma+b The reactant concentration in a zero-order reaction was 6.00x10 2 mol L-1 after 145 s and 3.00x10-2 mol L-1 after 370 s. What is the rate constant for this reaction? Order Integrated Rate Law Graph Slope Express your answer with the appropriate units VS View Available Hint(s) s.tk Al 2 [Alp koth = 1 Value Units Submit Part B What was the initial reactant concentration for the reaction described in Part A? Express your answer with the appropriate units View Available Hint(s) AlValue UnitsExplanation / Answer
The integrated rate law for a zero order reaction is given as
[A]t = -kt + [A]0
where k is the zero order rate constant.
Part A
We have [A]t = 6.00*10-2 mol.L-1 when t = 145 s and [A]t = 3.00*10-2 mol L-1 when t = 370 s. Therefore,
(6.00*10-2 mol L-1) = -k*(145 s) + [A]0 ……(1)
(3.00*10-2 mol L-1) = -k*(370 s) + [A]0 ……(2)
(1) – (2) gives
(6.00*10-2 – 3.00*10-2) mol L-1 = -k*(145 – 370) s
====> 3.00*10-2 mol L-1 = -k*(-225 s)
====> k = (3.00*10-2 mol L-1)/(225 s) = 1.333*10-4 mol L-1 s-1 ? 1.3*10-4 mol L-1 s-1
Therefore, we have k0th = 1.3*10-4 mol L-1 s-1 (ans).
Part B
Put k0th = 1.3*10-4 mol L-1 s-1 in (1) and obtain
6.00*10-2 mol L-1 = -(1.3*10-4 mol L-1 s-1)*(145 s) + [A]0
====> 6.00*10-2 mol L-1 = -(0.01885 mol L-1) + [A]0
====> [A]0 = (6.00*10-2 + 0.01885) mol L-1 = 0.07885 mol L-1 = 7.885*10-2 mol L-1 ans).
Part C
The integrated rate law for a first order reaction is given as
ln [A]t = -kt + ln [A]0
where k is the first order rate constant.
We have [A]t = 7.20*10-2 mol.L-1 when t = 35 s and [A]t = 3.80*10-3 mol L-1 when t = 80 s. Therefore,
ln (7.20*10-2 mol L-1) = -k*(35 s) + ln [A]0 ……(1)
ln (7.20*10-2 mol L-1) = -k*(80 s) + ln [A]0 ……(2)
(1) – (2) gives
ln (7.20*10-2 mol L-1) - ln (7.20*10-2 mol L-1) = -k*(35 – 80) s
====> ln (7.20*10-2 mol L-1)/(3.80*10-3 mol L-1) = -k*(-45 s)
====> ln (18.9474) = k*(45 s)
====> 2.9417 = k*(45 s)
====> k = (2.9417)/(45 s) = 0.06537 s-1 ? 0.065 s-1
Therefore, k1st = 0.065 s-1 (ans)
Part D
The integrated rate law for a second order reaction is given as
1/[A]t = kt + 1/[A]0
where k is the first order rate constant.
We have [A]t = 0.810 mol.L-1 when t = 285 s and [A]t = 7.40*10-2 mol L-1 when t = 875 s. Therefore,
1/(0.810 mol L-1) = k*(285 s) + 1/[A]0 ……(1)
1/(7.40*10-2 mol L-1) = k*(875 s) + 1/[A]0 ……(2)
(1) – (2) gives
1/(0.810 mol L-1) – 1/(7.40*10-2 mol L-1) = k*(285 - 875) s
====> (1.2346 L mol-1) – (13.5135 L mol-1) = k*(-590 s)
====> -12.2789 L mol-1 = -590*k s
====> k = (12.2789 L mol-1)/(590 s) = 0.02081 L mol-1 s-1 ? 0.210 L mol-1 s-1
Therefore, k2nd = 0.021 L mol-1s-1 (ans)