Answer :
To determine the correct expression for the variation of the concentration of a reactant as a function of time, we need to understand the order of the reaction and the integrated rate laws for different orders of reactions.
For a reaction involving a single reactant [tex]\( A \)[/tex] breaking down, the rate laws are different depending on whether the reaction is first-order, second-order, etc.
1. Zero-order reactions have a rate that is constant and independent of the concentration of [tex]\( A \)[/tex].
Integrated rate law:
[tex]\[ [A]_t = [A]_0 - k \cdot t \][/tex]
Here, [tex]\( k \)[/tex] is the rate constant, [tex]\( [A]_t \)[/tex] is the concentration of [tex]\( A \)[/tex] at time [tex]\( t \)[/tex], and [tex]\( [A]_0 \)[/tex] is the initial concentration of [tex]\( A \)[/tex].
2. First-order reactions have a rate that is directly proportional to the concentration of [tex]\( A \)[/tex].
Integrated rate law:
[tex]\[ \ln([A]_t) = \ln([A]_0) - k \cdot t \quad \text{or} \quad \ln([A]_t/[A]_0) = - k \cdot t \][/tex]
Rearranged form:
[tex]\[ \ln([A]_t) - \ln([A]_0) = - k \cdot t \][/tex]
3. Second-order reactions have a rate proportional to the square of the concentration of [tex]\( A \)[/tex].
Integrated rate law:
[tex]\[ \frac{1}{[A]_t} = \frac{1}{[A]_0} + k \cdot t \][/tex]
Given the choices:
A) [tex]\(\frac{[A]_0}{[A]_t} = k \cdot t\)[/tex]
B) [tex]\(\ln[A]_t - \ln[A]_0 = k \cdot t\)[/tex]
C) [tex]\(\frac{1}{[A]_t} = \frac{1}{[A]_0} + k \cdot t\)[/tex]
D) [tex]\(\frac{1}{[A]_0} - \frac{1}{[A]_t} = k \cdot t\)[/tex]
E) None of these answers
From the analysis:
- Choice (A) does not correspond to the integrated rate law of a zero, first, or second-order reaction.
- Choice (B) should be modified to [tex]\(\ln[A]_t - \ln[A]_0 = - k \cdot t\)[/tex] for a first-order reaction.
- Choice (C) matches exactly with the integrated rate law for a second-order reaction.
- Choice (D) does not correspond to any standard order reaction integrated rate law.
Thus, the expression that correctly represents the variation of concentration as a function of time is:
[tex]\[ \boxed{C} \][/tex]
For a reaction involving a single reactant [tex]\( A \)[/tex] breaking down, the rate laws are different depending on whether the reaction is first-order, second-order, etc.
1. Zero-order reactions have a rate that is constant and independent of the concentration of [tex]\( A \)[/tex].
Integrated rate law:
[tex]\[ [A]_t = [A]_0 - k \cdot t \][/tex]
Here, [tex]\( k \)[/tex] is the rate constant, [tex]\( [A]_t \)[/tex] is the concentration of [tex]\( A \)[/tex] at time [tex]\( t \)[/tex], and [tex]\( [A]_0 \)[/tex] is the initial concentration of [tex]\( A \)[/tex].
2. First-order reactions have a rate that is directly proportional to the concentration of [tex]\( A \)[/tex].
Integrated rate law:
[tex]\[ \ln([A]_t) = \ln([A]_0) - k \cdot t \quad \text{or} \quad \ln([A]_t/[A]_0) = - k \cdot t \][/tex]
Rearranged form:
[tex]\[ \ln([A]_t) - \ln([A]_0) = - k \cdot t \][/tex]
3. Second-order reactions have a rate proportional to the square of the concentration of [tex]\( A \)[/tex].
Integrated rate law:
[tex]\[ \frac{1}{[A]_t} = \frac{1}{[A]_0} + k \cdot t \][/tex]
Given the choices:
A) [tex]\(\frac{[A]_0}{[A]_t} = k \cdot t\)[/tex]
B) [tex]\(\ln[A]_t - \ln[A]_0 = k \cdot t\)[/tex]
C) [tex]\(\frac{1}{[A]_t} = \frac{1}{[A]_0} + k \cdot t\)[/tex]
D) [tex]\(\frac{1}{[A]_0} - \frac{1}{[A]_t} = k \cdot t\)[/tex]
E) None of these answers
From the analysis:
- Choice (A) does not correspond to the integrated rate law of a zero, first, or second-order reaction.
- Choice (B) should be modified to [tex]\(\ln[A]_t - \ln[A]_0 = - k \cdot t\)[/tex] for a first-order reaction.
- Choice (C) matches exactly with the integrated rate law for a second-order reaction.
- Choice (D) does not correspond to any standard order reaction integrated rate law.
Thus, the expression that correctly represents the variation of concentration as a function of time is:
[tex]\[ \boxed{C} \][/tex]
