CHEM102 Lecture Notes - Lecture 3: Rate Equation, Bayerischer Rundfunk, Torr
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In integrated rate law, time is a variable. Solving this equation, we get the integrated, linearized first order reaction . Rate= - [a] (cid:2930) = d[a]d(cid:2930) =k[a] ln [a] = ln [a]0 kt or ln[]0[]=kt: k varies with temperature, temperature in the reactions are fixed to keep k constant. Linearized first order reaction: ln[a]=ln[a](cid:2868) kt ln [a]0 plot ln[a] in the y axis and t on the axis as follows . If the reaction really is a first order reaction, then a straight line must be attained for this plot with a negative slope. K= - slope = (cid:889). (cid:885)(cid:882) x (cid:883)(cid:882) (cid:2872) (cid:2869) [h2o2]0 = e(cid:2868). (cid:2876)(cid:2872)(cid:2868)=(cid:884). (cid:885)(cid:883)(cid:888)=(cid:884). (cid:885)(cid:884) m y-intercept = +0. 840 ln [h2o2]0 = 0. 840 y-intercept helps to . Check if the initial concentration found from graph is the same as the experiment. Get the i(cid:374)itial (cid:272)o(cid:374)(cid:272)e(cid:374)tratio(cid:374) i(cid:374) (cid:272)ase it (cid:272)ould(cid:374)(cid:859)t (cid:271)e o(cid:271)tai(cid:374)ed duri(cid:374)g e(cid:454)peri(cid:373)e(cid:374)t (cid:449)hile (cid:373)i(cid:454)i(cid:374)g.