Questions: Step 2. Graphing the data. The graph of the data that gives a straight line indicates the correct integrated rate law. Data from a single experiment are graphed three ways. The graph of [A] vs t is curve sloping downward. The graph of ln [A] vs t is a line with a negative slope, and the graph of 1 /[A] vs t is a curve sloping upward. Based on these descriptions, what rate law applies to this reaction? A The reaction is first order and the rate law is Rate =k[A] B The reaction is second order and the rate law is Rate =k[A]^2 C The reaction is zeroth order and the rate law is Rate =k

Step 2. Graphing the data. The graph of the data that gives a straight line indicates the correct integrated rate law. Data from a single experiment are graphed three ways. The graph of [A] vs t is curve sloping downward. The graph of ln [A] vs t is a line with a negative slope, and the graph of 1 /[A] vs t is a curve sloping upward. Based on these descriptions, what rate law applies to this reaction? A The reaction is first order and the rate law is Rate =k[A] B The reaction is second order and the rate law is Rate =k[A]^2 C The reaction is zeroth order and the rate law is Rate =k
Transcript text: 9:47 Question 1.b of 5 Submit Step 2. Graphing the data. The graph of the data that gives a straight line indicates the correct integrated rate law. Data from a single experiment are graphed three ways. The graph of $[A]$ vs $t$ is curve sloping downward. The graph of $\ln [A]$ vs $t$ is a line with a negative slope, and the graph of $1 /[\mathrm{A}]$ vs $t$ is a curve sloping upward. Based on these descriptions, what rate law applies to this reaction? A The reaction is first order and the rate law is Rate $=k[A]$ B The reaction is second order and the rate law is Rate $=k[A]^{2}$ C The reaction is zeroth order and the rate law is Rate $=\mathrm{k}$ Tap here for additional resources
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Solution

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Solution Steps

Step 1: Analyze the Graphs

The problem provides three different graphs for the concentration of reactant \([A]\) over time \(t\):

  • \([A]\) vs \(t\) is a curve sloping downward.
  • \(\ln [A]\) vs \(t\) is a straight line with a negative slope.
  • \(1/[A]\) vs \(t\) is a curve sloping upward.
Step 2: Determine the Order of Reaction

For a reaction to be first order, the plot of \(\ln [A]\) vs \(t\) should be a straight line with a negative slope. This matches the description given in the problem.

  • Zeroth Order: \([A]\) vs \(t\) would be a straight line.
  • First Order: \(\ln [A]\) vs \(t\) is a straight line with a negative slope.
  • Second Order: \(1/[A]\) vs \(t\) would be a straight line.

Since the graph of \(\ln [A]\) vs \(t\) is a straight line with a negative slope, the reaction is first order.

Final Answer

The reaction is first order and the rate law is \(\text{Rate} = k[A]\).

\[ \boxed{\text{A}} \]

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