(10 points, total) A substrate (S) consumption rate is given by the following eq
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Question
(10 points, total) A substrate (S) consumption rate is given by the following equation: dS/dt=-k(S)2
(ia) (5 points) Develop an equation to calculate final substrate concentration (Sf) by assuming the following initial and final conditions: Sf = concentration @ time t; S0 = cell concentration @ time t = 0.
(ib) Provide units (or dimensions) for the kinetic constant “k”.
(ic) What is the order the reaction?
(ii) (5 points) Assume a bacterium is growing steadily in a closed bottle with nutrients. It doubles in the bottle every minute. At 11: 00 a.m. there was only one bacterium and at noon the bottle is full with bacteria. At what time the bottle was half full? Show your calculations
Explanation / Answer
(ia) (5 points) Develop an equation to calculate final substrate concentration (Sf) by assuming the following initial and final conditions: Sf = concentration @ time t; S0 = cell concentration @ time t = 0.
2S P
rate of reaction = : dS/dt=-k(S)2
-d[S]0/dt = k[S]^2
Now integrate both side:
1/[S]=kt+c
when t=0, [S]=[S]0
C=1/[S]0
1/[S]=kt+c
1/[S]=kt+1/[S]0
kt = 1/[S] - 1/[S]0
(ib) Provide units (or dimensions) for the kinetic constant “k”.
This is a second order reaction and for a second-order reaction, the rate of reaction is directly proportional to the square of the concentration of one of the reactants.
Differential Rate Law: r = k [S]2
The rate constant, k, has units of L mole-1 sec-1.
(ic) What is the order the reaction?
This is a second order reaction and for a second-order reaction, the rate of reaction is directly proportional to the square of the concentration of one of the reactants.
Differential Rate Law: r = k [S]2
(ii) (5 points) Assume a bacterium is growing steadily in a closed bottle with nutrients. It doubles in the bottle every minute. At 11: 00 a.m. there was only one bacterium and at noon the bottle is full with bacteria. At what time the bottle was half full? Show your calculations
Second order rate constant is given by the following equation:
kt = 1/[S]t - 1/[S]0
It doubles in the bottle every minute = 60sec
dS/dt=-k(S)2
(1-2)/60 = -k(1.0^2)
k= 0.0167
First calculate the concentration after 1.0 h = 3600 sec.
k=- 8. 33*10^-3
kt = 1/[S]t - 1/[S]0
kt + 1/[S]0 = 1/[S]t
0.0167*3600 +1/1.0 = 1/[S]t
1/[S]t=61.12
Now calculate the time the bottle was half full means 1/[S]t=61.12/2= 30.56
kt = 1/[S]t - 1/[S]0
kt + 1/[S]0 = 1/[S]t
0.0167*t +1/1.0 = 30.56
0.0167*t = 29.56
t= 1770 sec or 29.50 min
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