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?
C_{9}H_{8}O_{4} + NaOH -------> C_{9}H_{7}O_{4}Na + H_{2}O
Concentration of C_{9}H_{8}O_{4} = 5x10^{-4}M
NaOH = 7.1x10^{-5}M
**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...
3)Solution:
From the given data,
C_{9}H_{8}O_{4} + NaoH
C_{9}H_{7}O_{4} Na +
H_{2}O
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.
3. Determine a value of the pseudo-order rate constant, kobs. What is the half-life of the...
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