Step 1: Understanding the Concept:
This question asks about the integrated rate law for a zero-order reaction and its graphical representation. We need to know the relationship between reactant concentration and time for a zero-order process.
Step 2: Key Formula or Approach:
The rate law for a zero-order reaction (A \( \rightarrow \) Products) is:
\[ \text{Rate} = -\frac{d[A]}{dt} = k[A]^0 = k \]
where [A] is the concentration of the reactant and k is the rate constant.
To find the relationship between concentration and time, we need to integrate this rate law.
\[ -\frac{d[A]}{dt} = k \implies d[A] = -k \, dt \]
Integrating both sides from time \( t=0 \) (concentration \( [A]_0 \)) to time t (concentration \( [A]_t \)):
\[ \int_{[A]_0}^{[A]_t} d[A] = \int_0^t -k \, dt \]
\[ [A]_t - [A]_0 = -kt \]
Rearranging this gives the integrated rate law:
\[ [A]_t = -kt + [A]_0 \]
Step 3: Detailed Explanation:
The integrated rate law for a zero-order reaction is:
\[ [A]_t = -kt + [A]_0 \]
This equation is in the form of a straight line, \( y = mx + c \), where:
- \( y = [A]_t \) (the concentration of the reactant at time t)
- \( x = t \) (time)
- \( m = -k \) (the slope of the line)
- \( c = [A]_0 \) (the y-intercept, which is the initial concentration)
Therefore, a plot of the reactant concentration ([Reactant]) on the y-axis against time (t) on the x-axis will be a straight line with a slope equal to \( -k \).
Step 4: Final Answer:
The correct statement is that a plot of [Reactant] against time is a straight line with a slope equal to -k.