A first order reaction is 50% completed in 1.26 × 10^14 s. How much time would it take for 100% completion?
Answer: 100 min
Explanation: The integrated rate law for a first-order reactionlooks like this
ln(AA0)=−k⋅t , where#
A - the concentration at a given time t
A0 - the initial concentration
k - the rate constant, usually expressed in s−1for first-order reactions
Now, I'll assume that you're not familiar with the equation that describes the half-life of a first-order reaction.
As you know, the half-life of a chemical reaction tells you how much time is needed for the concentration of a reactant to reach half of its initial value.
Simply put, the time needed for 50% of a reaction to be completed will represent that reaction's half-life.
In your case, you know that your first-order reaction is 50% completed after 30 minutes. This means that after 30 minutes, you will have
A=12⋅A0→ half of the initial concentration remains after one half-life
Plug this into the above equation and solve for k, the rate constant
ln(12⋅A0A0)=−k⋅30 min
This is equivalent to
−k⋅30 min=ln(12)
−k⋅30 min==0ln(1)−ln(2)
k=ln(2)30 min=2.31⋅10−2min−1
Now that you know the rate constant for this reaction, use the same equation again, only this time use
A=110⋅A0→ equivalent to 90%completion
and solve for t, the time needed to get here
ln(110⋅A0A0)=−2.31⋅10−2min−1⋅t
This is equivalent to
t=−ln(10)−2.31⋅10−2min−1=99.68 min
100 min
Explanation:The integrated rate law for a first-order reactionlooks like this
ln(AA0)=−k⋅t , where#
A - the concentration at a given time t
A0 - the initial concentration
k - the rate constant, usually expressed in s−1for first-order reactions
Now, I'll assume that you're not familiar with the equation that describes the half-life of a first-order reaction.
As you know, the half-life of a chemical reaction tells you how much time is needed for the concentration of a reactant to reach half of its initial value.
Simply put, the time needed for 50% of a reaction to be completed will represent that reaction's half-life.
In your case, you know that your first-order reaction is 50% completed after 30 minutes. This means that after 30 minutes, you will have
A=12⋅A0→ half of the initial concentration remains after one half-life
Plug this into the above equation and solve for k, the rate constant
ln(12⋅A0A0)=−k⋅30 min
This is equivalent to
−k⋅30 min=ln(12)
−k⋅30 min==0ln(1)−ln(2)
k=ln(2)30 min=2.31⋅10−2min−1
Now that you know the rate constant for this reaction, use the same equation again, only this time use
A=110⋅A0→ equivalent to 90%completion
and solve for t, the time needed to get here
ln(110⋅A0A0)=−2.31⋅10−2min−1⋅t
This is equivalent to
t=−ln(10)−2.31⋅10−2min−1=99.68 min
