If the half of a reactant in a 1st order reaction is 25 minutes, then what is the extent of reaction after 2 minutes?

Question

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Answer 1

The half-life of a first order reaction is given by the equation:

t_1/2 = ln(2) / k

Solving for k, the reaction rate constant:

k = ln(2) / t_1/2

k = ln(2) / 25min = 0.693 / 25min

k = 0.028 / min

The integrated rate law for a first order reaction is:

[A] = [A]_o * e^-kt

Now knowing the rate constant k =0.028 / min, after 2 minutes...

[A] = [A]_o * e^{-(0.028 / min * 2 min)}

[A] = 0.946 * [A]_o

...there is approximately 94.6% of the original amount of reactant A remaining. The reaction has proceeded 100% - 94.6% = 5.4%.

So, 5.4% of the starting amount of reactant A has reacted.