For the reaction $ A \rightarrow $ products,
The reaction was started with 2.5 mol L\(^{-1}\) of A.
For the reaction $ A \rightarrow $ products,
The reaction was started with 2.5 mol L\(^{-1}\) of A.
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Correct Answer: 2435
Approach Solution - 1
From the graph, we know that \( t_{1/2} \) is proportional to \( [A] \).
The slope is given as 76.92. Thus, using the equation for zero-order reaction: \[ t_{1/2} = \frac{A_0}{2K} \quad \text{where} \quad \text{slope} = \frac{1}{2K} = 76.92 \] Thus, \[ K = \frac{1}{2 \times 76.92} = \frac{1}{153.84} \] Now, applying the formula for zero-order reaction: \[ [A] = - Kt + A_0 \] \[ [A] = - \frac{1}{2 \times 76.92} \times 10 + 2.5 = 2.435 \, \text{mol/L} \] Thus, the concentration of A at 10 minutes is \( 2435 \times 10^{-3} \, \text{mol/L} \).
Approach Solution -2
The problem provides a graph of the half-life (\(t_{1/2}\)) versus the initial concentration (\([A]_0\)) for the reaction \(A \rightarrow \text{products}\). Based on this graph and the given initial concentration, we need to calculate the concentration of A after 10 minutes.
Concept Used:
The relationship between the half-life (\(t_{1/2}\)) of a reaction and the initial concentration of the reactant (\([A]_0\)) depends on the order of the reaction (\(n\)). The general relation is \(t_{1/2} \propto [A]_0^{1-n}\).
- For a zero-order reaction (\(n=0\)): \(t_{1/2} \propto [A]_0\). The formula is \(t_{1/2} = \frac{[A]_0}{2k}\).
- For a first-order reaction (\(n=1\)): \(t_{1/2}\) is independent of \([A]_0\). The formula is \(t_{1/2} = \frac{\ln 2}{k}\).
- For a second-order reaction (\(n=2\)): \(t_{1/2} \propto \frac{1}{[A]_0}\). The formula is \(t_{1/2} = \frac{1}{k[A]_0}\).
The integrated rate law for a zero-order reaction is given by:
\[ [A]_t = [A]_0 - kt \]
where \([A]_t\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant.
Step-by-Step Solution:
Step 1: Determine the order of the reaction from the given graph.
The graph shows a straight line passing through the origin for a plot of \(t_{1/2}\) (y-axis) versus \([A]_0\) (x-axis). This indicates a direct proportionality between the half-life and the initial concentration: \(t_{1/2} \propto [A]_0\). This relationship is characteristic of a zero-order reaction.
Step 2: Relate the slope of the graph to the rate constant (\(k\)).
For a zero-order reaction, the half-life is given by the formula \(t_{1/2} = \frac{[A]_0}{2k}\). This can be written as \(t_{1/2} = \left(\frac{1}{2k}\right) [A]_0\). Comparing this to the equation of a straight line, \(y = mx\), where \(y = t_{1/2}\) and \(x = [A]_0\), the slope (\(m\)) is equal to \(\frac{1}{2k}\). We are given that the slope is 76.92.
\[ \text{Slope} = \frac{1}{2k} = 76.92 \]
Step 3: Calculate the rate constant (\(k\)).
From the relationship in Step 2, we can solve for \(k\). The units of the slope are \(\frac{\text{min}}{\text{mol L}^{-1}}\).
\[ 2k = \frac{1}{76.92} \] \[ k = \frac{1}{2 \times 76.92} = \frac{1}{153.84} \text{ mol L}^{-1} \text{min}^{-1} \]
Step 4: Use the integrated rate law for a zero-order reaction to find the concentration of A at \(t = 10\) minutes.
The integrated rate law is \([A]_t = [A]_0 - kt\). We are given:
- Initial concentration, \([A]_0 = 2.5 \text{ mol L}^{-1}\)
- Time, \(t = 10 \text{ min}\)
- Rate constant, \(k = \frac{1}{153.84} \text{ mol L}^{-1} \text{min}^{-1}\)
Substituting these values into the equation:
\[ [A]_{10} = 2.5 \text{ mol L}^{-1} - \left(\frac{1}{153.84} \text{ mol L}^{-1} \text{min}^{-1}\right) \times 10 \text{ min} \]
Step 5: Compute the final concentration.
\[ [A]_{10} = 2.5 - \frac{10}{153.84} \] \[ [A]_{10} \approx 2.5 - 0.06500 \] \[ [A]_{10} \approx 2.435 \text{ mol L}^{-1} \]
The question asks for the answer in the form of ____ \(\times 10^{-3} \text{ mol L}^{-1}\). We convert our result to this format:
\[ 2.435 \text{ mol L}^{-1} = 2435 \times 10^{-3} \text{ mol L}^{-1} \]
Rounding to the nearest integer, the concentration of A at 10 minutes is 2435 \(\times 10^{-3} \text{ mol L}^{-1}\).
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