If 75% of a first order reaction was completed
in 90 minutes, 60% of the same reaction would
be completed in approximately (in minutes)
_______.
(Take : log 2 = 0.30; log 2.5 = 0.40)
Answer (integer)
60
Solution
t<sub>75%</sub> = 90 min = 2 × t<sub>1/2</sub>
<br><br>$\Rightarrow$ t<sub>1/2</sub> = 45 min
<br><br>Rate constant, K = ${{0.693} \over {45}}$ min<sup>-1</sup>
<br><br>Time for completion of 60% of the reaction,
<br><br>t<sub>60%</sub> = ${{2.303} \over K}\log {{10} \over 4}$
<br><br>= ${{2.303 \times 45} \over {0.693}}\log 2.5$
<br><br>= 60 min
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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