Unit 5: Topic 3 - Concentration Changes over Flashcards
Since a plot of concentration vs. time is linear, the reaction is in zeroth order. If a plot of ln[A] vs. t is linear, the reaction is first order, and if a plot of 1/[A] vs. t is linear, the reaction is second order.
If a chemical reaction is first-order with respect to reactant A, derive the rate law and provide a graph depicting the change in concentration with respect to time.
If a chemical reaction is second-order with respect to reactant A, plot a graph representing concentration vs time and determine which plot is linear.
Consider the reaction A –> B. The initial concentration of reactant A is 0.800M.
After 10 seconds, the concentration of reactant A is .600M, and after 20 seconds, the concentration of reactant A is .400M. Determine the order of the reaction with respect to A and calculate the rate constant.
If we plot these three values, we note that the graph of [A] vs. t is linear. Therefore, we know that this is a zeroth-order reaction. Since the reaction equation is [A]t - [A]0 = -kt, we can solve for k. Using the concentration at 10 seconds, we have .600M - .800M = -(10)k, so that k = 0.02 M/s
Consider the reaction A –> B. The initial concentration of reactant A is 0.500M.
After 10 seconds, the concentration of reactant A is 0.200M, and after 20 seconds, the concentration of reactant A is 0.125M. Determine the order of the reaction with respect to A and calculate the rate constant.
If we plot these three values, obviously the graph of [A] vs. t is not linear. Therefore, we check to see if ln[A] vs. t is linear or 1/[A] vs. t is linear. The values of ln[A] are given by ln0.5 = -0.693, ln0.2 = -1.609, and ln0.125 = -2.079: a quick sanity check shows these values are not linear. The values of 1/[A] are 2, 5, and 8: these are linear. Therefore, since 1/[A] vs. t is linear, this is a second-order reaction. The reaction equation is 1/[A]t - 1/[A]t0 = kt, we can use the concentration at 10 seconds: 1/(0.200) - 1/(0.500) = 10k, so k = 0.30. Recalling units, note that Rate = k[A]^2. The rate has units of M/s, and [A] has units of M, so k should have units M^-1 s^-1. Therefore, k = 0.30 M^-1 s^-1.
Explain why the half-life for a first-order reaction is constant.
Recall the first-order rate equation ln[A]t = -kt + ln[A]0. We can rearrange this equation as ln[A]t - ln[A]0 = -kt, and combining the logarithms give ln([A]t/[A]0) = -kt. A half-life is defined as the time required for a substance to decay to half of its original concentration, so we should have [A]t = 1/2 * [A]0. Thus the fraction [A]t/[A]0 = 1/2, so ln1/2 = -.693 = -kt. Then rearrange so that t_{1/2} = .693/k is constant. Therefore, due to the equation of concentration for a first-order reaction, the half-life is constant and independent of a starting concentration.
5.0g of substance A decomposes to 2.0g after 30 minutes. What is the half-life of substance A assuming first-order kinetics?
Assuming first-order kinetics, we know that the half-life is constant. Therefore, we have t_(1/2) = 0.693/k. From the first-order equation, ln([A]t/[A]0) = -kt, we have ln(2/5) = -30k, or k = ln(5/2)/30 because mass ratio equals concentration ratio. Substituting this into the equation for half-life, we have t_(1/2) = 30 * 0.693/ln(5/2) ≈ 22.69 minutes. In general, given the time t and concentrations [A]0 and [A]t, the half-life is t_(1/2) = t * ln2 / ln([A]0/[A]t).
Iodine-131 decays to Xenon-131 with a half-life of 8 days. If a container holds 16g of iodine-131, how much remains after 16 days? Explain why this is an example of a first-order reaction.
Half-life is constant in a first-order reaction and is not constant in zero or second-order reactions. Therefore, radioactive decay, which has constant half-lives, resembles first-order reactions. After 16 days, two half-life periods have passed, so the amount in the container should be half twice, for a total of 16g * 1/4 = 4g of iodine-131 after 16 days.
