Question

# 3. Determine a value of the pseudo-order rate constant, kobs. What is the half-life of the...

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 )

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.

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