Chem 112 - /04/2013& Chemical Kinetics Key Topics o...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 14 09/04/2013 ° Chemical Kinetics Key Topics o Chemical Kinetics o Rates of Reactions o Concentration Change over time o Activation Energy o Reaction Mechanisms ° ° Kinetics Studying the rate of chemical reactions Simply, how fast a reaction proceeds o Rate = (∆) Change in concentration / (∆) Time The rate is most helpful in determining a reaction mechanism o Shows the individual steps of how a reaction proceeds There are 2 types of reaction rates that we will consider: o Average Rate over (∆) Time o Instantaneous Rate Example: Write the “Rate Law” for this decomposition reaction 2N2O5  NO2 + O2 Rate = - ∆(N2O5) / ∆T
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
° 14.2a: Average Rate: Rates change over the course of a reaction because? o Concentration is changing The reaction takes time. Not all of reactants are converted to products all at once. o 2N2O5  4NO2 + O2 Rate (mol / L * T) = - ∆(N2O5) / ∆T  Slope equation between 2 points Using Reaction Stoichiometry – The Relative Average Rate o The rate of the reaction compared to reactants and products o Simply, the rate of decrease of reactants and the rate of increase of products aA + bB  cC + dD Rate of decrease of reactants: Rate = -(1/a) * ∆(A) / ∆(T) = -(1/b) * ∆(B) / ∆(T) Rate of appearance of products Rate = (1/c) * ∆(C) / ∆(T) = (1/d) * ∆(D) / ∆(T) (Reactant) is decreasing, so NEGATIVE ° 14.2b: Instantaneous Rate The rate of a reaction at a specific point in time o This can be found by drawing a tangent line at the point of interest At time t = 0, the rate is known as the initial rate o Rate = -∆[N2O5] / ∆T ° 14.3: Effect of concentration on rate
Image of page 2
N2O5  NO2 + (½)O2 o In this case, doubling concentration doubles the rate Rate = k[N2O5] K = rate constant For the general equation, Rate = k[A]^x[B]^y The rate of the reaction is proportional to the concentration of the reactants raised to some power ° 14.3: Reaction Order Helps determine how the concentration effects raction rate o Must be determined experimentally, not from stoichiometric coefficients 2NO + Cl2  2NOCl If we double the concentration of [NO], rate quadruples. Rate is proportional to [NO]^2, we say it is second order in this reactant If we double the concentration of [Cl2], rate doubles. Rate is proportional to [Cl2], we say it is first order in this reactant. So, for this example, rate = k[NO]^2[Cl2] We say the overall order is third, which is the total sum of the exponents. If the reaction order is zero in some reactant, then the rate of that reaction does not depend on the concentration of that reaction o Reaction mechanisms and intermediates From experimental data, the rate law for this reaction is: K[NO2]^2 We say the reaction is 2 nd order in NO2 and zero for CO.
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern