Chem 104 1st Edition Lecture 27 Outline of Last Lecture I. Review of important equationsII. Effect of temperature on ratea. Suvante ArrheniusIII. Activation energy and activated complexOutline of Current LectureI. Two point formCurrent LectureI. The two point form is the form of solving he prefers, but will allow you to solve however works best for you.The two point form:−EaR=ln(k2k2)(1T2−1T1)This will be provided on the test.Example:For the reaction NO2+CO→CO2+NO has a rate constant of 2.57 M-1s-1 at 701 K, and a rate constant of 567 M-1s-1 at 895 K. Find the activation energy in kj/mol.−EaR=ln(k2k2)(1T2−1T1)−Ea8.314 J=ln(5672.57)(1895−1701)−Ea8.314 J=5.3964−3.0921∗1 0−4−Ea=1.45∗105J =145 kJ /molExample:The rate of reaction doubles every 10°C rise in temperature. Calculate the activation energy for such reaction.Given:T2=310 KT1= 300 KRearrange the equation to look like this:ln(k2k1)=EaR(1T2−1T1)ln(k2k1)=Ea8.314(1310−1300)Now if the rate of reaction doubles every 10°C, then K2 = 2K2ln(2 k2k1)=Ea8.314(1310−1300)ln(2)=Ea8.314(1310−1300)Ea=5.359∗1 0−4Jmol=53.6kJmolExample: Molecule 1 dissociates at 625 K with first order k=0.271s-1. How long will it take for 60% reactant to decompose?Given: [I2] =5.0 MAfter some time [I2] =3.0 MOnce again rearrange to look like this:I¿¿2¿t¿¿¿¿¿ln ¿I¿¿2¿t¿¿¿¿¿ln ¿Now you can rearrange this equation to look like this:I2¿0I2¿t=−kt +ln ¿ln ¿ln (2.5)=−0.271 t +ln (5.0)[I2]=0.40∗[I2]ln (.40)=−0.271tt =ln
View Full Document