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In a first order reaction, the concentration of the reactant, decreases from 0.8 M to 0.4 M in 15
minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is -           [AIEEE-2004]
  • a)
    30 minutes
  • b)
    15 minutes
  • c)
    7.5 minutes
  • d)
    60 minutes
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
In a first order reaction, the concentration of the reactant, decrease...
The correct answer is  Option A.
The concentration of the reactant decreases from 0.8M to 0.4M in 15 minutes, i.e. t1/2 = 15 minutes. Therefore, the concentration of the reactant will fall from 0.1M to 0.025M in two half lives. i.e.    2t1/2 = 2 × 15 = 30 minutes.
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Most Upvoted Answer
In a first order reaction, the concentration of the reactant, decrease...
If the concentration changes from 0.8 M to 0.4 M in 15 minutes that means the concentration is getting halved in 15 minuter (or) the half life of the reaction is 15 minutes. 
Now when the concentration changes from 0.1 M to 0.025 M it means the concentration is becoming one-fourth the initial value which is possible when the time period is equal to 2 half lives (for one half life it becomes 1/2 and for the other half it becomes 1/4). So, the time taken is 30 minutes.
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Community Answer
In a first order reaction, the concentration of the reactant, decrease...
Since in 1st case concentration is getting halved so we can apply formula
k= 0.693/t (1/2) where the value of t 1/2 is 15 min.
Now, time taken for concentration to change is
t = 2.303/k log a/(a - x)
on putting values we get
t = 2.303 * 15/0.693 log (0.1/0.025)
t = 29.9 min ~ 30 min
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In a first order reaction, the concentration of the reactant, decreases from 0.8 M to 0.4 M in 15minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is - [AIEEE-2004]a)30 minutesb)15 minutesc)7.5 minutesd)60 minutesCorrect answer is option 'A'. Can you explain this answer?
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