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If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately
- a)45 minutes
- b)60 minutes
- c)40 minutes
- d)50 minutes
Correct answer is option 'A'. Can you explain this answer?
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If 60% of a first order reaction was completed in 60 minutes, 50% of t...
First Order Reaction and Half-Life
A first-order reaction is a chemical reaction in which the rate of the reaction is proportional to the concentration of a single reactant. This means that the rate of the reaction decreases as the concentration of the reactant decreases over time.
The half-life of a first-order reaction is the time required for the concentration of the reactant to decrease by half. The half-life of a first-order reaction is independent of the initial concentration of the reactant and is a constant characteristic of the reaction.
Solution
Given that 60% of a first-order reaction was completed in 60 minutes, we can use the half-life equation to determine the half-life of the reaction.
ln(0.60) = -kt
where k is the rate constant of the reaction, t is the time elapsed, and ln is the natural logarithm.
Solving for k, we get:
k = -ln(0.60)/60
k ≈ 0.0115 min^-1
Using the half-life equation, we can find the time required for the concentration of the reactant to decrease by half:
t1/2 = ln(2)/k
t1/2 ≈ 60.3 minutes
Therefore, if 60% of the first-order reaction was completed in 60 minutes, then 50% of the reaction would be completed in approximately:
t = t1/2 × ln(0.50)/ln(0.60)
t ≈ 45 minutes
Thus, the correct answer is option A, 45 minutes.
A first-order reaction is a chemical reaction in which the rate of the reaction is proportional to the concentration of a single reactant. This means that the rate of the reaction decreases as the concentration of the reactant decreases over time.
The half-life of a first-order reaction is the time required for the concentration of the reactant to decrease by half. The half-life of a first-order reaction is independent of the initial concentration of the reactant and is a constant characteristic of the reaction.
Solution
Given that 60% of a first-order reaction was completed in 60 minutes, we can use the half-life equation to determine the half-life of the reaction.
ln(0.60) = -kt
where k is the rate constant of the reaction, t is the time elapsed, and ln is the natural logarithm.
Solving for k, we get:
k = -ln(0.60)/60
k ≈ 0.0115 min^-1
Using the half-life equation, we can find the time required for the concentration of the reactant to decrease by half:
t1/2 = ln(2)/k
t1/2 ≈ 60.3 minutes
Therefore, if 60% of the first-order reaction was completed in 60 minutes, then 50% of the reaction would be completed in approximately:
t = t1/2 × ln(0.50)/ln(0.60)
t ≈ 45 minutes
Thus, the correct answer is option A, 45 minutes.
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Community Answer
If 60% of a first order reaction was completed in 60 minutes, 50% of t...
Find K for 60%
k=1/60 * {ln100/40} = 0.01527
(reaction is 60%completed so remaining is 40%)
t(50%) =1/K * {ln100/50} = 45 minute
k=1/60 * {ln100/40} = 0.01527
(reaction is 60%completed so remaining is 40%)
t(50%) =1/K * {ln100/50} = 45 minute
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If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximatelya)45 minutesb)60 minutesc)40 minutesd)50 minutesCorrect answer is option 'A'. Can you explain this answer?
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