CHEM 113 1st EditionLecture 3Outline of Last Lecture I. Expressing the reaction ratea. 3 ways to describe rateII. The rate law and its componentsa. Reaction ordersOutline of Current Lecture II. Solving for the rate lawa. Determining reaction ordersb. Rate constantIII. Integrated rate lawCurrent Lecture16.3: The rate law and its components cont. - Solving for the rate law-o Here is a general plan that we will use to determine the reaction order, rate constant and eventually the entire rate equation. Some problems given in class were only one step of the following plan, however we were also given problems that required us to go through all the steps to get the answer. 1). Given: Series of plots of concentration vs. time- From this we determine the slope of the tangent at t0, giving us the initial rate for each plot. This was discussed the in previous lecture. 2). Given/Found: Initial rates- We then compare the initial rates when the concentration of A changes and B is held constant and vice versa. This gives us the reaction orders. This step is explained in more detail in the next section. 3). Given/Found: concentrations and reaction orders- We then can plug them into the rate law and solve for our rate constant k. 4). Given/Found: Initial rates, rate constant, orders, and concentrations ofall reactantsThese 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.- We then substitute these values into the rate law rate=k[A]^m[B]^n- Determining Reaction Orderso This means determining the values of m and no This can only be done by conducting an experiment and collecting data points. We will most likely be given a data chart to use and manipulate to solve for thesevalues.o The following uses Table 16.2 on page 637 as an example. To find the reaction order of A, we first see where B is constant and A changes. In this case, this occurs in the first two rows of the table. We then use the following equation- The rates are the values found in column labeled Initial Rate.- Like when calculating slope, the second recorded data point goes on top. Rate 2 and [A]2 are values found from the second experimental trial. Rate 2 = [A]2 Rate 1 = [A]1 After dividing we obtain value=value^m- For example, if we got 2.00=(2.00)^m, this would mean it was a first order reaction.- However sometimes this may be mathematically difficult to compute. In these cases you can use o m=log a/log b where a=b^m You then repeat using the other reactant as the changing value and the first reactant as the constant.- Rate constanto Once we know the reaction orders (m and n). we can solve for k rate=k[A]^m[B]^no Simply substitute concentration and rates into the rate equation.o However, the units for k change depending on the overall reaction order Overall reaction order: 0- Units: mol/(L*s) Overall reaction order: 1- Units: 1/s or s^-1 Overall reaction order: 2- Units: L/(mol*s) Overall reaction order: 3- Units: L^2/(mol^2 * s)16.4: Integrated rate laws: concentration changes over time- Now we can predict concentration in time.- Integrated rate law: includes time as a variable.o They differ depending on the order. First-order rate equation- ln ([A]0/[A]t) = -kt- [A]0= Initial concentration and [A]t= concentration at specific time. Second-order rate equation- (1/[A]t)-(1/[A]0) = kt Zero-order rate equation- [A]t-[A]0 = -kt- Using these laws graphicallyo Each of these laws can be rearranged to fit the format of y=mx+b, therefore they all make a linear line when graphed. o This is helpful because you may be given a graph and be able to use it to identify values of your formulao y=mx+b form of the integrated rate laws First order: ln[A]t= -kt + ln[A]0 Second order: (1/[A]t)= kt + (1/[A]0)- Note that when graphed this forms a positive slope while the other two orders make a negative slope. Zero order: [A]t= -kt +[A]0o You may be asked to look at a set of data points and identify what order it is. To do this you have to plot the points. If they make a linear line then it is 0th order. If it does not, find the natural log of all the points. If this makes a linear line then it is 1st order. If it does not you must find the inverse of all the points. If that makes a linear line then it is 2nd order. Because it will most likely be experimental data that you are looking at, the points may not make a perfect linear line. Therefore you have to find the order that is closest to a linear