Solveeit Logo

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{\text{mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 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.

Explanation

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 + BC + D{\text{A + B}} \to {\text{C + D}} , according to rate law expression, Rate = k[A]x[B]y{\text{Rate = k[A}}{{\text{]}}^{\text{x}}}{{\text{[B]}}^{\text{y}}} , where x + y{\text{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.
APA \to P , for a reaction in which reactant A gives product P, the rate can be written asRate = - d[A]dt{\text{Rate = - }}\dfrac{{{\text{d[A]}}}}{{{\text{dt}}}} , and now let us consider it as a zero-order reaction.
We know that at a time t=0t = 0 , the concentration of A is A0{A_0} whereas, at a time t, the concentration of A is At{A_t}.
Since this is a zero-order reaction, the rate law expression can be written as
Rate=d[At]dt=k[At]0Rate = - \dfrac{{d[{A_t}]}}{{dt}} = k{[{A_t}]^0} , k is the rate constant for this reaction. {since rate is independent of reactant concentration in a zero-order reaction}
d[At]dt=k- \dfrac{{d[{A_t}]}}{{dt}} = k Because anything raised to zero is 1 and applying this to the above equation.
d[At]t=kdt- d{[{A_t}]_t} = kdt (rearranging)
So now let us integrate both sides, we get
dAt=kdt- \int {d{A_t} = k\int {dt} } (Since k is a constant it can be taken out from the integration)
[A]t=kt+c- {[A]_t} = kt + c
As we already discussed above when time is zero sec t=0t = 0s, the concentration of A is [A]0{[A]_0}. So we can find the integration constant, [A]0=k(0)+c - {[A]_0} = k(0) + c
So we can write it as [A]0=c - {[A]_0} = c
[A]t=kt[A]0- {[A]_t} = kt - {[A]_0}
On rearranging, the rate constant ,k=[A]0[A]ttk = \dfrac{{{{[A]}_0} - {{[A]}_t}}}{t}
So from the above equation, we can find the unit of the rate constant, k=concentrationtimek = \dfrac{{concentration}}{{time}}
k=mol/Lsk = \dfrac{{mol/L}}{s} 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 molL1s1mol{L^{ - 1}}{s^{ - 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{t_{1/2}} , [A]=[A]02[A] = \dfrac{{{{[A]}_0}}}{2}
Thus, we can write the rate equation for zero-order at half-life.
k=[A]0[A]02t1/2k = \dfrac{{{{[A]}_0} - \dfrac{{{{[A]}_0}}}{2}}}{{{t_{1/2}}}}
k=[A]02t1/2\Rightarrow k = \dfrac{{{{[A]}_0}}}{{2{t_{1/2}}}}
t1/2=[A]02k\Rightarrow {t_{1/2}} = \dfrac{{{{[A]}_0}}}{{2k}}
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{{\text{n}}^{{\text{th}}}} order reaction can be found out easily using the formula mol1 - n Ln - 1 s - 1{\text{mo}}{{\text{l}}^{{\text{1 - n}}}}{\text{ }}{{\text{L}}^{{\text{n - 1}}}}{\text{ }}{{\text{s}}^{{\text{ - 1}}}} . The rate of reaction depends on the temperature, concentration, pressure, presence of catalyst except in zero-order which is independent of concentration.