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Integrated Rate Law quiz #1 Flashcards

Integrated Rate Law quiz #1
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  • Which of the following equations represents the integrated rate law for a first-order reaction?
    The integrated rate law for a first-order reaction is: ln([A]_t) = -kt + ln([A]_0), where [A]_t is the concentration of reactant at time t, [A]_0 is the initial concentration, k is the rate constant, and t is time.
  • What is the integrated rate law equation for a zero-order reaction?
    The equation is [A]_t = -kt + [A]_0, where [A]_t is the concentration at time t, [A]_0 is the initial concentration, k is the rate constant, and t is time.
  • How do the units of the rate constant (k) differ for zero, first, and second order reactions?
    For zero order, k has units of molarity/time; for first order, k has units of 1/time; for second order, k has units of 1/(molarity·time).
  • What graphical feature indicates a zero-order reaction when plotting concentration versus time?
    A zero-order reaction produces a straight line with a negative slope when concentration is plotted against time.
  • Which type of reaction is always first order with respect to its integrated rate law?
    Radioactive decay processes always follow a first-order integrated rate law.
  • What is the significance of a plot of ln(concentration) versus time in reaction kinetics?
    A linear plot of ln(concentration) versus time indicates a first-order reaction.
  • How can you identify a second-order reaction from a graph?
    A second-order reaction is identified by a straight line with a positive slope when 1/[A] is plotted versus time.
  • What does the slope represent in the integrated rate law graph for a zero-order reaction?
    The slope represents -k, the negative of the rate constant, indicating a decrease in concentration over time.
  • In the second-order integrated rate law, what does the y-intercept correspond to?
    The y-intercept corresponds to 1/[A]_0, the inverse of the initial concentration.
  • Why is the slope positive in the integrated rate law graph for a second-order reaction?
    The slope is positive because the equation is 1/[A]_t = kt + 1/[A]_0, so as time increases, 1/[A]_t increases linearly.