If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately
ln(No/N) = kt
Solve for k using 100 for No and 40 for N(i.e., if it was 60% completed and you started with 100, you would have 40 left.
Then ln(No/N) = kt again, use k from above and solve for t.
To determine the time required for 50% completion of the same reaction, we need to use the concept of half-life in first order reactions.
The half-life of a first order reaction is the time it takes for the concentration of the reactant to decrease by half. The formula to calculate the half-life of a first order reaction is:
t(1/2) = (0.693 / k)
where t(1/2) is the half-life and k is the rate constant of the reaction.
In this case, we are given that 60% of the reaction is completed in 60 minutes. This means that the remaining 40% will be completed in the next half-life.
So, let's first calculate the rate constant (k) using the given information:
k = (0.693 / t(1/2))
Since we know that 60% of the reaction is completed in 60 minutes, we can substitute the values into the equation solving for k:
k = (0.693 / 60)
k ≈ 0.01155 min^(-1)
Now, we can use the rate constant to calculate the half-life of the reaction:
t(1/2) = (0.693 / k)
t(1/2) ≈ (0.693 / 0.01155)
t(1/2) ≈ 60 minutes
So, the half-life of the reaction is approximately 60 minutes. Therefore, if 60% of the reaction is completed in the first 60 minutes, 50% of the reaction will be completed in the next half-life, which is also approximately 60 minutes.