To calculate the time required for the completion of 90% of a reaction with first-order kinetics, we can use the equation:
t = (0.693/k) * (1 - e^(-kt))
Where:
- t is the time required for the completion of a certain percentage of the reaction (in this case, 90%)
- k is the rate constant for the reaction
- e is the base of the natural logarithm (approximately equal to 2.71828)
Given that the half-life (T1/2) is 44.1 minutes, we can find the rate constant (k) using the formula for a first-order reaction:
k = (0.693) / T1/2
Substituting the value of T1/2 into the equation, we get:
k = (0.693) / 44.1
k ≈ 0.0157 min^-1
Now, we can calculate the time required for 90% completion using the equation mentioned earlier:
t = (0.693/k) * (1 - e^(-kt))
Substituting the values:
t = (0.693 / 0.0157) * (1 - e^(-0.0157t))
To solve for t, we can rearrange the equation:
(1 - e^(-0.0157t)) = (0.1 * 0.0157 / 0.693)
Simplifying further:
e^(-0.0157t) = 0.0143
Taking the natural logarithm (ln) of both sides:
-0.0157t = ln(0.0143)
Solving for t:
t = ln(0.0143) / -0.0157
t ≈ 87.8 minutes
Therefore, it would take approximately 87.8 minutes for 90% of the reaction to complete, given the half-life of 44.1 minutes.