3. Determine a value of the pseudo-order rate constant, kobs. What is the half-life of the reaction under the conditions of the experiment? For how many half-lives was the reaction allowed to run?
C9H8O4 + NaOH -------> C9H7O4Na + H2O
Concentration of C9H8O4 = 5x10-4M
NaOH = 7.1x10-5M
**Second order reaction in respect to acetylsalicylic acid.
**The rate law may be written as, rate = k[asp]m[OH– ]n and
The pseudo-rate law is rate = kobs[asp]m Where kobs = k[OH– ]n or ln(kobs) = ln(k[OH– ]n )
If more information is needed please let me know in comments...
3)Solution:
From the given data,
C9H8O4 + NaoH
C9H7O4 Na +
H2O
Rate = –d[asp]/dt = k [asp]2
Rate = –d[OH-]/dt = k [OH-]
Rate = d[aspNa]/dt = k [aspNa]
And
[asp] = 5 ×10-4 M
[NaOH] or [OH-] = 7.1 ×10-5 M
For, a complete reaction rate = k
[asp]2 [OH-]
For pseudo-order reaction , rate = kobs[asp]2
-------------------- (1)
Where kobs = k [OH-]
So, the overall order of the reaction is 2.
Therefore, for second order reaction
Rate = –d[asp]/dt = kobs [asp]2
and for second order reaction kobs = (1/t) [(1/a-x) -
(1/a)] or kobs = (1/t) {x / [a (a-x) ]}
--------------------(2)
Where a and x are intial and final concentrations of asp respectively
and kobs = 1 / t1/2 . a
or t1/2 = 1 /
kobs . a ---------------------------
(3)
So, from the above equations 1, 2 and 3, to solve the above
problem we need to have at least one of these values of rate or
rate constant k or time t ( half-life t1/2 ) of the
reaction.
Then you can find a solution to the given problem.
3. Determine a value of the pseudo-order rate constant, kobs. What is the half-life of the...
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