Question
Question: In a zero-order reaction:- A. The rate constant has the unit \({\text{mol }}{{\text{L}}^{{\text{ -...
In a zero-order reaction:-
A. The rate constant has the unit mol L - 1s - 1
B. The rate is independent of the concentration of the reactants.
C. The half-life depends on the concentration of the reactants.
D. The rate is independent of the temperature of the reaction.
Solution
-The rate of reaction is the speed at which a reaction takes place. The reaction rate is expressed in terms of concentration. We know the rate law expression. It states that the rate of a reaction can be expressed in terms of the reactant concentration raised to a power that is determined experimentally. The power can be 0, 1, 2, or 3, etc. If the power of the reactant concentration is zero, it is called a zero-order reaction. For a reaction A + B→C + D , according to rate law expression, Rate = k[A]x[B]y , where x + y gives the order of the reaction. (Anything raised to the power zero is 1)
Complete step by step answer:
From the above-mentioned explanation, we can conclude that for a zero-order reaction, the order is zero. This means that the rate does not depend on the concentration of the reactants in a zero-order reaction. Thus, option B is correct.
These reactions are independent of reactant concentration.
A→P , for a reaction in which reactant A gives product P, the rate can be written asRate = - dtd[A] , and now let us consider it as a zero-order reaction.
We know that at a time t=0 , the concentration of A is A0 whereas, at a time t, the concentration of A is At.
Since this is a zero-order reaction, the rate law expression can be written as
Rate=−dtd[At]=k[At]0 , k is the rate constant for this reaction. {since rate is independent of reactant concentration in a zero-order reaction}
−dtd[At]=k Because anything raised to zero is 1 and applying this to the above equation.
−d[At]t=kdt (rearranging)
So now let us integrate both sides, we get
−∫dAt=k∫dt (Since k is a constant it can be taken out from the integration)
−[A]t=kt+c
As we already discussed above when time is zero sec t=0s, the concentration of A is [A]0. So we can find the integration constant, −[A]0=k(0)+c
So we can write it as −[A]0=c
−[A]t=kt−[A]0
On rearranging, the rate constant ,k=t[A]0−[A]t
So from the above equation, we can find the unit of the rate constant, k=timeconcentration
k=smol/L as we know the unit of concentration is mol per litre and that of time is sec.
Unit of the rate constant, k for a zero-order reaction is molL−1s−1
Thus, we could say that option A is correct.
Half-life is the time when the concentration of the reactant is reduced to half its initial concentration.
So, for a zero-order reaction, at half-life t1/2 , [A]=2[A]0
Thus, we can write the rate equation for zero-order at half-life.
k=t1/2[A]0−2[A]0
⇒k=2t1/2[A]0
⇒t1/2=2k[A]0
Therefore half-life depends on the concentration of the reactant in zero-order reaction. Therefore option C is correct.
We know that the rate of a reaction depends on its temperature. Thus, option D is incorrect.
So, the correct answer is Option A,B,C.
Note: If the power of the concentration of the reactant is 1, it is a first-order reaction and 2 for second-order reaction and goes on. The unit of the rate constant of nth order reaction can be found out easily using the formula mol1 - n Ln - 1 s - 1 . The rate of reaction depends on the temperature, concentration, pressure, presence of catalyst except in zero-order which is independent of concentration.