In a certain first order reaction, 99.9% of the reaction is complete in 1355s. Calculate the time taken for 40 % of the reaction to be complete.
answer in brief
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Answer:
Answer: Time required to complete 99.9 % reaction is 149.5 mins.
Explanation: For first order reactions, rate is written as:
k=\frac{2.303}{t}log\left(\frac{a}{a-x}\right)k=
t
2.303
log(
a−x
a
)
where, k = rate constant
t = time required
a = product formed
a - x = reactant left in the reaction
When the reaction is 75 % completed
a = 100
a - x = 100 - 75 = 25
Putting the values in rate equation, we get
k=\frac{2.303}{30\times 60}log\left(\frac{100}{25}\right)k=
30×60
2.303
log(
25
100
)
k=7.703\times 10^{-4}sec^{-1}k=7.703×10
−4
sec
−1
As the rate constant for the reaction remains same. Now, when the reaction is 99.9% completed
a = 100
a - x = 100 - 99.9 = 0.1
Putting the values in the above rate equation, we get
7.703\times 10^{-4}=\frac{2.303}{t}log\left(\frac{100}{0.1}\right)7.703×10
−4
=
t
2.303
log(
0.1
100
)
t = 8969.92 sec or (Conversion factor: 1 min = 60 sec)
t = 149.5 mins
100 per sent step by step explation