Question

What is the half-life of a first-order reaction with a rate constant of $ 7.8 \times {10^{ - 4}}{s^{ - 1}} $ ?

Answer 1

Hint :A reaction's half-life is that time it takes for a reactant to exceed half of its initial concentration or strain. The half-life of a first-order reaction is concentration independent and remains constant along with passage of time. Here since the rate constant that is denoted by $ k $ of this particular reaction is mentioned we can easily substitute in the formula of reactions of first order for finding half-life or $ {T_{1/2}} $ .

Complete Step By Step Answer:
For half-life of first order reaction;
 $ \Rightarrow {t_{1/2}} = \dfrac{{\ln (2)}}{k} $
As we begin to solve this numerical, we can keep the steps in mind: First write down the given values, then write the formula again, see which all values can be substituted in the formula from the given values and finally after substituting the values we get the final answer which is half-life here.
Given values;
They mentioned it is first order.
Rate $ k = 7.8 \times {10^{ - 4}}{s^{ - 1}} $
Now the formula is written as;
 $ \Rightarrow {t_{1/2}} = \dfrac{{\ln (2)}}{k} $
We see that the numerator is constant, so we only need denominator and denominator is the rate constant that is given in the question.
So now we substitute the rate constant into the reaction;
 $ \Rightarrow {t_{1/2}} = \dfrac{{\ln (2)}}{{7.8 \times {{10}^{ - 4}}{s^{ - 1}}}} $
 $ \Rightarrow {t_{1/2}} = 889s $
So the final result gives that the half-life of first-order reaction of the given rate is $ \Rightarrow {t_{1/2}} = 889s $ .

Note :
The half-life of an element is extremely useful and has different applications. It is different for different reactions though. According to the change in the reaction’s order, the formula of half-life is different. Sometimes when we are finding the half-life formula we might need a reaction’s order, rate or initial concentration.