Part A The reactant concentration in a zero-order reaction was 8.00x10-2 mol L-1
ID: 575096 • Letter: P
Question
Part A The reactant concentration in a zero-order reaction was 8.00x10-2 mol L-1 after 200 s and 1.00x102 mol L- after 315 s. What is the rate constant for this reaction? Express your answer with the appropriate units. View Available Hint(s) 2 IA MI koth Submit Previous Answers Request Answer X Incorrect; Try Again; 6 attempts remaining 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) [A]o= 1 Value UnitsExplanation / Answer
Part A
The integrated rate law for a zero order reaction is given as
[A] = -k*t + [A]0
where [A] is the concentration of the reactant at time t and [A]0 is the initial reactant concentration; k is the zero order rate constant.
When t = 200 s, we have [A] = 8.00*10-2 mol L-1 and when t = 315 s, we have [A] = 1.00*10-2 mol L-1. Plug in these values and obtain
8.00*10-2 mol L-1 = -k*(200 s) + [A]0 ……(1)
1.00*10-2 mol L-1 = -k*(315 s) + [A]0 ……(2)
Subtract (2) from (1) and obtain
(8.00*10-2 mol L-1) – (1.00*10-2 mol L-1) = -k*(200 – 315) s
====> 7.00*10-2 mol L-1 = 115*k s
====> k = 6.0869*10-4 mol L-1 s-1 6.087*10-4 mol L-1 s-1 6.087*10-4 M/s (ans).
Part B
Put the value of k in (1) above and obtain
8.00*10-2 mol L-1 = -(6.087*10-4 mol L-1s-1)*(200 s) + [A]0
====> 8.00*10-2 mol L-1 = -0.12174 mol L-1 + [A]0
====> [A]0 = 8.00*10-2 mol L-1 + 0.12174 mol L-1 = 0.20174 mol L-1 0.202 M (ans).
Part C
The integrated rate law for a first order reaction is given as
ln [A] = -k*t + ln [A]0
where [A] is the concentration of the reactant at time t and [A]0 is the initial reactant concentration; k is the zero order rate constant.
When t = 10 s, we have [A] = 9.90*10-2 mol L-1 and when t = 95 s, we have [A] = 2.40*10-3 mol L-1. Plug in these values and obtain
ln (9.90*10-2) = -k*(10 s) + ln [A]0 ……(1)
ln (2.40*10-3) = -k*(95 s) + ln [A]0 …….(2)
Subtract (2) from (1) and obtain
ln (9.90*10-2) – ln (2.40*10-3) = -k*(10 – 95) s
====> ln (9.90*10-2/2.40*10-3) = 85*k s
====> 3.71965 = 85*k s
====> k = 3.71965/85 s-1 = 0.04376 s-1 0.0438 s-1 (ans)
Part D
The integrated rate law for a second order reaction is given as
1/[A] = k*t + 1/[A]0
where [A] is the concentration of the reactant at time t and [A]0 is the initial reactant concentration; k is the zero order rate constant.
When t = 275 s, we have [A] = 0.890 mol L-1 and when t = 850 s, we have [A] = 4.40*10-2 mol L-1. Plug in these values and obtain
1/(0.890 mol L-1) = k*(275 s) + 1/[A]0 ……(1)
1/(4.40*10-2 mol L-1) = k*(850 s) + 1/[A]0 …….(2)
Subtract (2) from (1) and obtain
1/(0.890 mol L-1) – 1/(4.40*10-2 mol L-1) = k*(275 – 850) s
====> (1.12359 L/mol) – (22.72727 L/mol) = -575*k s
====> -21.60368 L/mol = -575*k s
====> k = 21.60368/575 L mol-1s-1 = 0.0375716 M-1s-1 0.0376 M-1s-1 (ans)