Find rate constant by plugging numbers into integrated rate law Point out that

# Find rate constant by plugging numbers into

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Find rate constant by plugging numbers into integrated rate law. Point out that often k is found in one equation and used in another equation. 39mod. A first-order reaction is 38.5% complete in 480 s. (a) Calculate the rate constant. ln[C] t = -kt + ln[C] o ln[0.615 * ] t = (-k x 480) + ln k = 1.0128 x 10 -3 s -1 * Note that 38.5% complete means that at time “t” 61.5% of the reactant is still present. Chem 162-2016 Chapter 20 Electrochemistry 63 Determination of the rate constant from the half-life formula . The half life of a first order reaction is 2.64s. What is the rate constant? t 1/2 = 0.693/k 2.64s = 0.693/k k = 0.2625s -1 Chem 162-2016 Chapter 20 Electrochemistry 64 CHEM-2010 FINAL EXAM + ANSWERS CHAPTER 12 - KINETICS RATES,RATE CONSTANTS, REACTION ORDERS, HALF-LIVES CONCEPTS INTEGRATED RATE LAW Determination of the rate constant from the slope of the integrated-rate-law line. Since a plot of ln[A] vs time gives a straight line, then the reaction is first order. ln[A] t = -kt + ln[A] o Y = mX + b slope = m = -k slope = (Y 2 -Y 1 )/(X 2 -X 1 ) = (-3.60 –(-1.47))/(1195-0) = -1.78 x 10 -3 s -1 -k = slope k = -slope = 1.78 x 10 -3 s -1 B = Answer (X 2 Y 2 ) (X 1 Y 1 ) ln concentration Chem 162-2016 Chapter 20 Electrochemistry 65 Finding other things after reaction order and k are found 2 Chem 162-2014 Exam I H&P Chapter 13A – Chemical Kinetics RATES,RATE CONSTANTS, REACTION ORDERS, HALF-LIVES CALCULATIONS The second-order reaction, 2Mn(CO) 5 Mn 2 (CO) 10 has a rate constant equal to 3.0 × 10 9 M -1 s -1 at 25°C. If the initial concentration of Mn(CO) 5 is 1.0 × 10 -5 M, how long will it take for 90.% of the reactant to disappear? A. 3.3 × 10 -16 s B. 3.7 × 10 -15 s C. 3.0 × 10 -4 s D. 3.0 × 10 3 s E. 1.0 × 10 -3 s ET Note: Percentages only work in place of concentrations for 1 st order reactions. A second order reaction, given two concentrations and time, calls for the integrated rate law: 1/[C] t = kt + 1/[C] o k = 3.0 × 10 9 M -1 s -1 [C] o = 1.0 x 10 -5 M 90% of the reactant disappearing means that there is 10% remaining at time t. (1/(0.10 x (1.0 x 10 -5 ))) = ((3.0 x 10 9 ) x t) + (1/(1.0 x 10 -5 )) t = 3.0 x 10 -4 s Chem 162-2016 Chapter 20 Electrochemistry 66 ARRHENIUS EQUATION FORMULAS transparency Rate equation: Rate = k[X] m [Y] n “k” is the rate constant The “k” in the rate law is the same as the “k” in the Arrhenius equation. Arrhenius equation: k = Ae (-Ea/RT) ln k = -E a /RT + ln A Note: y = mx + b; y=ln(k); m=(-Ea/R); x=(1/T); b=ln(A) A = frequency factor = combination of steric factor and collisional frequency E a = energy of activation ln k 2 - ln k 1 = (-E a /RT 2 ) - (-E a /RT 1 ) ln (k 2 /k 1 ) = -((E a2 - E a1 )/R) x (1/T) This last formula probably will not be on the exam formula sheet so should be memorized. IMPORTANT “RATE” FACTORS Rate = k[X] m [Y] n • Concentration of reactants Reaction order • k (Arrhenius rate constant) IMPORTANT “k” FACTORS ln k = (-E a /R) x 1/T + ln A • Collision frequency (part of A) • Steric factor (also called orientation factor [part of A]) • Energy of activation Catalyst • Temperature ln k = (-E a /R) x 1/T + ln A ln (k 2 /k 1 ) = -(E a /R) x [(1/T 2 ) - (1/T 1 )] Chem 162-2016 Chapter 20 Electrochemistry 67 THE ARRHENIUS EQUATION -- COLLISION THEORY Rate = k[X] m [Y] n “k” is the rate constant The “k” in the rate law is the same as the “k” in the Arrhenius equation. Arrhenius equation: k = Ae (-Ea/RT) ln k = (-E a /RT) + ln A E a = energy of activation R = 8.314Jdeg -1 mol -1  #### You've reached the end of your free preview.

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