JEE MAIN - Chemistry (2020 - 4th September Morning Slot - No. 9)
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)
(Take : log 2 = 0.30; log 2.5 = 0.40)
Answer
60
Explanation
t75% = 90 min = 2 × t1/2
$$ \Rightarrow $$ t1/2 = 45 min
Rate constant, K = $${{0.693} \over {45}}$$ min-1
Time for completion of 60% of the reaction,
t60% = $${{2.303} \over K}\log {{10} \over 4}$$
= $${{2.303 \times 45} \over {0.693}}\log 2.5$$
= 60 min
$$ \Rightarrow $$ t1/2 = 45 min
Rate constant, K = $${{0.693} \over {45}}$$ min-1
Time for completion of 60% of the reaction,
t60% = $${{2.303} \over K}\log {{10} \over 4}$$
= $${{2.303 \times 45} \over {0.693}}\log 2.5$$
= 60 min
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