HalfLifeActivationEnergy

# HalfLifeActivationEnergy - Reaction Half-Life What is...

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What is Half-Life? Reaction Half-Life The time it takes for the concentration of reactants to decrease by one-half [A] t = k(T)t + [A] 0 -1 -1 [A] 0 0 th Order 1 st Order 2 nd Order Rate Law Integrated Rate Law Half Life Rate=k(T) Rate=k(T)[A] Rate=k(T)[A] 2 [A] t = -k(T)t + [A] 0 ln[A] t = -k(T)t + ln[A] 0 t ! = [A] 0 2k(T) t ! = k(T) t ! = k(T) 1 Reaction Laws Overview 0.693 ln 2 Example SO 2 Cl 2 (g) ! SO 2 (g) + Cl 2 (g) The disappearance of SO 2 Cl 2 is monitored experimentally 0.035 80 0.081 60 0.187 40 0.433 20 1.000 0 [SO 2 Cl 2 ] (M) Time (min) 100 0.015 What should you do with this data?

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What is the order of the reaction in SO 2 Cl 2 ? Look at the plots of concentration vs. time 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 20 40 60 80 100 Time (min) [SO 2 Cl 2 ] The reaction is first order in SO 2 Cl 2 - Slope = 0.042 min -1 = k(T) [SOCl 2 ] Time (min) What is the half life of the reaction? t 1/2 = ln 2 k(T) = 0.693 0.042 min -1 = 16.5 min After 16.5 min, half of the SOCl 2 is reacted 0 1.0M 16.5 0.5M 0.2M 33 49.5 0.125M 66 0.063M
What is the concentration of SO 2 Cl 2 at 74 minutes? Use the first order integrated rate law ln [A] = -k(T)t + ln[A] 0 (-0.042 min -1 ) (74 min) + ln(1.00M) = -3.108 [A] = ln [A] = e -3.108 = 0.044 M at 74 mins. t = ln [A] - ln[A] 0 -k(T) = ln (0.263M) - ln(1.000M) -0.042 min -1 = 31.8 min Use the first order integrated rate law ln [A] = -k(T)t + ln[A] 0 At what time is the concentration of SO 2 Cl 2 at 0.263M?

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