CHEM 1332 1st Edition Lecture 7 Outline of Last Lecture - Different types of solubility - Temperature effects of solubility- Henry’s Law- Concentration effects on rate of reactionOutline of Current Lecture - Rate Law- Order- Half-lifeCurrent Lecture- To determine the rate law the formula is rate=k [reactant 1]x[reactant 2]y - An important thing to note is that you cannot just look at a reaction and decide the order- An example is rate=k NO2xH2y- First you determine values of x and y the order- Then you determine the k if you canExperimentNO2mol /LH2mol / LRate mol/L*sec1 0.1 0.2 62 0.2 0.2 123 0.2 0.4 48- As the reaction takes place you need to measure something like color- In this experiments the first change from the first and second is the increase in NO2 which is doubled the change in rate is doubled by this change- From experiment 2 and 3 there is an increase in H2 which is doubled the rate is quadrupled - From the observations we can predict that x=1 and y=2- The new equation is rate=k NO21H22-rate=k NO21H22-6=k(0.1)(0.2)2These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- k=1500- If we take a look at just one reactant it is a decomposition reaction - Taking a closer look at this as a first order reaction the rate law is rate=k[A]- If we try to find the units of k they vary from each reaction different orders equal different k - First order the units of k are 1/time - The integrated law is ln A0−ln At=kt- The equation of the half-life of a first order is ln 2=kt 1/2- This equation is independent of concentration- In a second order reaction the rate law is rate=k A2- The units of k is L/mol*sec- The integrated rate law is 1At−1A0=kt- The half-life is 1At=kt 1/2Key TermsHalf-life- the time taken to use up half of the starting
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