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Question
The time for half change in a first order decomposition of a substance A is 60 seconds. Calculate the rate constant. How much of A will be left after 180 seconds?
Solution
1. Order of a reaction = 1
`"t"_(1/2)` = 60 seconds,
k = ?
We know that, k = `0.6932/"t"_(1/2)`
k = `2.303/60` = 0.01155 s−1
2. [A0] = 100%
t = 180 s
k = 0.01155 seconds−1
[A] = ?
For the first order reaction k = `2.303/"t" log (["A"_0])/(["A"])`
0.01155 = `2.303/180 log (100/(["A"]))`
`(0.01155 xx 180)/2.303 = log (100/(["A"]))`
0.9207 = log 100 – log [A]
log [A] = log 100 – 0.9207
log [A] = 2 – 0.9207
log[A] = 1.0973
[A] = antilog of (1.0973)
[A] = 12.5%
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