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Concentration Natural Log of Time (s) of N 2 O 5 (mol/L) N 2 O 5 (mol/L)

Integrated Rate Laws 1. Mathematical expression focusing on the relationship between time and concentration in a reaction. 2. Applies only to a specific temperature and set of conditions. 3. Different forms of the equation for different orders of reaction.

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Concentration Natural Log of Time (s) of N 2 O 5 (mol/L) N 2 O 5 (mol/L)

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  1. Integrated Rate Laws1. Mathematical expression focusing on the relationship between time and concentration in a reaction. 2. Applies only to a specific temperature and set of conditions.3. Different forms of the equation for different orders of reaction

  2. Integrated Rate LawsFirst Order ReactionsFor reaction: A  product (s) Initial Rate = -  [A] = k[A] tIntegrated rate law ln[A]f = - ktf + ln[A]ot = t

  3. Integrated rate law ln[A]f = - ktf + ln[A]oif the reaction is first order, the plot of ln[A] vs tis a straight line where y = ln[A] and x = t ln[A]f = - ktf + ln[A]oy = a x + b

  4. Concentration Natural Log of Time (s) of N2O5 (mol/L) N2O5 (mol/L) 0.1000 - 2.303 0 0.0707 -2.249 50 0.0500 -2.996 100 0.0250 -3.689 200 0.0125 -4.382 300 0.00625 -5.075 400 Integrated Rate Law ln [N2O5]f – ln [N2O5]0 = - 0.00665 s-1 t Differential Rate Law rate = 0.00665 s-1 [N2O5]1

  5. Integrated Rate Law ln[N2O5]f – ln [N2O5]0 = - 0.00665 s-1 t What would be concentration of dinitrogen pentoxide after 150 sec if the initial concentration was 0.1 M ln [Af] – ln [0.1] = - 0.00665 s-1 (150 s) ln [Af] = - 3.3 [Af] = 0.0369 M

  6. Integrated Rate Law ln[N2O5]f – ln [N2O5]0 = - 0.00665 s-1 t How long would it take for the concentration of dinitrogen pentoxide to decrease to 0.003 M if the initial concentration was 0.1 M ln [0.003] – ln [0.1] = - 0.00665 s-1 t - 3.51-2.3 = t - 0.00665 s-1 527 s = t

  7. Integrated Rate Law ln[N2O5]f – ln [N2O5]0 = - 0.00665 s-1 t How long would it take for the concentration of dinitrogen pentoxide to decrease by 50% ln [0.05] – ln [.1] = - 0.00665 s-1 t -0.693 = t - 0.00665 s-1 104 s = t

  8. Example Problem Given the following rate law for the reaction: X2Y  2X + Y rate [X2Y/min] = 0.600 min -1 [X2Y] What would be the integrated rate law for the reaction? ln [X2Y]f – ln [X2Y]0 = - 0.600 min -1 t

  9. Example Problem Given the following rate law for the reaction: X2Y  2X + Y rate [X2Y/min] = 0.600 min -1 [X2Y] What would be the concentration of X2Y after 10 min if the initial concentration was 0.500 M? ln [X2Y]f – ln [X2Y]0 = - 0.600 min -1 t ln [X2Y]f – ln [0.500]0 = - 0.600 min -1 (10 min)

  10. Example Problem Given the following rate law for the reaction: X2Y  2X + Y rate [X2Y/min] = 0.600 min -1 [X2Y] What is the half-life of the reaction? ln [X2Y]f – ln [X2Y]0 = - 0.600 min -1 t ln [0.500]f – ln [1.00]0 = - 0.600 min -1 t

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