Question

3. Determine a value of the pseudo-order rate constant, k_obs. 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 = k_obs[asp]^m Where k_obs = k[OH^– ]ⁿ or ln(k_obs) = ln(k[OH^– ]ⁿ )

If more information is needed please let me know in comments...

Answer 1

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 = k_obs[asp]² -------------------- (1)

                                                    Where k_obs = k [OH^-]

So, the overall order of the reaction is 2.

Therefore, for second order reaction Rate = –d[asp]/dt = k_obs [asp]²    
and for second order reaction k_obs = (1/t) [(1/a-x) - (1/a)] or k_obs = (1/t) {x / [a (a-x) ]} --------------------(2)

Where a and x are intial and final concentrations of asp respectively

and k_obs = 1 / t_1/2 . a or   t_1/2  = 1 / k_obs . 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 t_1/2 ) of the reaction.
Then you can find a solution to the given problem.