Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
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]
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.