Measurement of the Activation Energy of a Reaction (Arrhenius equation). Calculate: activation energy, E a. Standard techniques of rearranging equations. Solution - CHAPTER 1 CHEMISTRY THE STUDY OF CHANGE Problem Categories Biological 1. Unformatted text preview: CHAPTER 1. CHEMISTRY: THE.
Rate = k[A]^x[B]^y if [A]^x[B]^y is kept constant for two reactions then Rate1/Rate2 = k1/k2 as k = Ae^-Ea/RT then Rate1/Rate2 = (Ae^-Ea/RT1)/(Ae^-Ea/RT2) A cancels out Rate1/Rate2 = (e^-Ea/RT1)/(e^-Ea/RT2) from expts 1 and 2 Rate1/Rate2 = 0.49934 thus 0.49934 = (e^-Ea/RT1)/(e^-Ea/RT2) take logs(e) ln 0.49934 = Ea/RT2 - Ea/RT1 = Ea/R x (1/T2-1/T1) ln 0. Access Serial Port Linux. 49934 x 8.314 = Ea [ (T1 - T2)/T1T2] (-0.69447 x 8.314) x T1T2/(T1-T2) = Ea (-5.7738 x 101099) /(-10) = Ea Ea = 58372 J Ea = 58.372 kJ better check my sums!! (Original post by charco) deep breath. Rate = k[A]^x[B]^y if [A]^x[B]^y is kept constant for two reactions then Rate1/Rate2 = k1/k2 as k = Ae^-Ea/RT then Rate1/Rate2 = (Ae^-Ea/RT1)/(Ae^-Ea/RT2) A cancels out Rate1/Rate2 = (e^-Ea/RT1)/(e^-Ea/RT2) from expts 1 and 2 Rate1/Rate2 = 0.49934 thus 0.49934 = (e^-Ea/RT1)/(e^-Ea/RT2) take logs(e) ln 0.49934 = Ea/RT2 - Ea/RT1 = Ea/R x (1/T2-1/T1) ln 0.49934 x 8.314 = Ea [ (T1 - T2)/T1T2] (-0.69447 x 8.314) x T1T2/(T1-T2) = Ea (-5.7738 x 101099) /(-10) = Ea Ea = 58372 J Ea = 58.372 kJ better check my sums!! Wow, i really appreciate that. I am going to have to decipher it lol.
This equation is derived from the Arrhenius equation: k= Ae^-Ea/RT take logs (either natural logs or base 10) natural ogs give you: lnk = lnA - Ea/RT base 10 logs give: log(10)k = log(10)A -Ea/2.303RT in either case you can plot graphs as they both have the form y = mx + c ( a straight line graph) OK s for your equation, log(10)k = log(10)A -Ea/2.303RT so, rearranging gives: log(10)k = -Ea/2.303RT + log(10)A (similar to y = mx + c) a plot of log(10) k (the y values) against -1/2.303RT (the x values) gives the Ea as the gradient of the line and log(10) A as the intercept with the y axis. Your graph is probably upside down because you have your axis the wrong way round.
A(g) + B(g) --->C(g) The rate law for the above reaction is: -d[A]/dt = k[A][B] The rate constant is 3.66×10^-5 L mol-1 s-1 at 348.0°C and 3.64×10^-2 L mol-1 s-1 at 545.0°C. 1) Use the Arrhenius equation k = Ae-Ea/RT to Calculate Ea for this reaction. 2) Calculate the rate constant k at 423.0°C. A(g) + B(g) --->C(g) The rate law for the above reaction is: -d[A]/dt = k[A][B] The rate constant is 3.66×10^-5 L mol-1 s-1 at 348.0°C and 3.64×10^-2 L mol-1 s-1 at 545.0°C. 1) Use the Arrhenius equation k = Ae-Ea/RT to Calculate Ea for this reaction. 2) Calculate the rate constant k at 423.0°C for this reaction.