Question

The half-life period of a zero order reaction, A $\to $ product is given by:
[A] $\dfrac{{{\left[ A \right]}_{\circ }}}{k}$
[B] $\dfrac{0.693}{k}$
[C] $\dfrac{{{\left[ A \right]}_{\circ }}}{2k}$
[D] $\dfrac{2{{\left[ A \right]}_{\circ }}}{k}$

Answer 1

Hint: To find the answer to this question, firstly write the rate law equation for a zero order reaction and from there find out the integrated equation of the rate. Use the relation between the initial concentration and the concentration at half-life period to find the concentration in terms of initial concentration. Substitute that in the integrated rate law equation and rearrange to get the correct answer.

Complete step by step answer:
 Before answering this, let us understand what a half-life period and a zero order reaction is.
We know that order of a reaction is the relation between the rate of the reaction and the concentration of the reactants in it. It tells us how the reaction rate is affected by the concentration of the involved reactants.
The half-life period of a radioactive element is the time required for the substance to reduce to half of its initial value.

Now let us find the half-life period of a zero order reaction.
We can write the rate law equation for the zero-order reaction as-
     \[Rate=k{{\left[ A \right]}^{n}}\]
Now, for zero order, n=0. Therefore, we can write the differential form of the rate law equation as-
     \[Rate=-\dfrac{d{{\left[ A \right]}^{\circ }}}{dt}=k\]

Rearranging it we can write that-
     \[d\left[ A \right]=-kdt\]
Integrating both sides and solving for A, we get the integrated form of the rate law equation which is-
     \[\left[ A \right]={{\left[ A \right]}^{\circ }}-kt\]

Now, we know the half-life period is the time when the concentration of the substance becomes half of the initial concentration. Therefore, we can write that-
     \[\left[ A \right]=\dfrac{1}{2}{{\left[ A \right]}^{\circ }}\]
Now, putting this value of the concentration of A in the integrated rate law equation, we get-
     \[\dfrac{1}{2}{{\left[ A \right]}^{\circ }}={{\left[ A \right]}^{\circ }}-k{{t}_{{1}/{2}\;}}\]
Or, we can write that-
     \[\dfrac{{{\left[ A \right]}^{\circ }}}{2k}={{t}_{{1}/{2}\;}}\]
That is, the half-life of a zero order reaction is dependent upon the initial concentration as well as the rate concentration.
From the discussion we can see that half-life period of a zero-order reaction is $\dfrac{{{\left[ A \right]}_{\circ }}}{2k}$
So, the correct answer is “Option C”.

Note: We have discussed above that the rate of reaction is affected by the concentration of the reactants. If we have a zero order reaction, it means that the concentration of the reactants do not affect the rate of the reaction. Similarly, for a first order reaction, the concentration of one of the reactant species affects the reaction rate.
The rate of a reaction depends upon several other factors like the nature of the reactants, presence of catalyst, temperature at which the reaction is taking place and also on the presence of catalyst. Here, as the reaction is zero ordered so further addition of reactants to increase the concentration of the reactant will not affect the rate of the reaction. The reaction will proceed at the same, constant speed.