help with kinetics problem

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inaccensa

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15+ Year Member
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Step 1 A→ B + C (slow)
Step 2 B + D → E + F (fast)
Step 3 E + C → G (fast)

In this case, the first step in the reaction pathway is the
rate-determining step. Therefore, the overall rate of the reaction
must equal the rate of the first step, k1[A] where k1 is the
rate constant for the first step. (Rate constants of the different
steps are denoted by kx, where x is the step number.)
In some cases, it is desirable to measure the rate of a
reaction in relation to only one species. In a second-order
reaction, for instance, a large excess of one species is
included in the reaction vessel. Since a relatively small
amount of this large concentration is reacted, we assume
that the concentration essentially remains unchanged.
Such a reaction is called a pseudo first-order reaction. A
new rate constant, k', is established, equal to the product of
the rate constant of the original reaction, k, and the concentration
of the species in excess
. This approach is often
used to analyze enzyme activity

In a test of the rate of Step 3 of Reaction 1, a solution
is prepared containing a 0.1 M concentration of E and
a 50 M concentration of C. The rate is calculated after
the reaction has gone 50% to completion. By what
percent will the calculated rate differ from the true
rate if we treat the reaction as pseudo first-order?
A 0.02%
B. 0.05%
C. 0.1%
D. 0.2%

here is my approach

R = K (E) (C)
K(.1M *.5) (50-.5)---> K (.05)(49.95)--->2.4975K
R = K' (C)
K'(50*.5)----> k' (25)

The difference between the 2
k'(25)- k(2.4975) = 22.50k

22.50/2.4975=9
 
Last edited:
2nd order: k[E][C]

at halfway [E] = 50 - 0.05, [C] = 0.05
rate = k(50 - 0.05)(0.05)


pseudo 1st order: 50k[C]

at halfway [C] = 0.05
rate = 50k(0.05)

answer is C
 
2nd order: k[E][C]

at halfway [E] = 50 - 0.05, [C] = 0.05
rate = k(50 - 0.05)(0.05)


pseudo 1st order: 50k[C]

at halfway [C] = 0.05
rate = 50k(0.05)

answer is C

i understood what you did for the first part, why did you multiply the concentration of C in the psuedo by 0.05? Can you plzz explain
 
i understood what you did for the first part, why did you multiply the concentration of C in the psuedo by 0.05? Can you plzz explain

I'm not multiplying the concentration of C by 0.05, the concentration of C IS 0.05M.

The initial concentration of C is 0.1M. At the half way point, the concentration of C is reduced by half to 0.05M regardless of how we solve for the rate. (Notice how I set [C] = 0.05M for the second order side as well)
 
I'm not multiplying the concentration of C by 0.05, the concentration of C IS 0.05M.

The initial concentration of C is 0.1M. At the half way point, the concentration of C is reduced by half to 0.05M regardless of how we solve for the rate. (Notice how I set [C] = 0.05M for the second order side as well)

Actually the concentration of C in the q is 50M, I realized that you interchanged the #'s in part one of the solution. For part I it didn't matter, but for part II, its confusing
 
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