PowerPoint Presentation

Dr. HarrisLecture 18HW: Ch 17: 5, 11, 18, 23, 41, 50Ch 17: Kinetics Pt 1Reactions RatesChemical kinetics is the area of chemistry that investigates how fast reactions occur

Different reactions proceed with different rates

The rate of a reaction depends on several factors, including:reactant concentrationtemperaturecatalystssurface area

Today, we will focus exclusively on the relationship between reaction rates and reactant concentrationIntroLets take the reaction: A ---> B.

This reaction tells us that as A is consumed, B is formed at an equal rate. We can express this mathematically in terms of changing concentrations by:

Imagine we have 10 moles of A in 1 L of solution. If we can freeze time for an instant, such that the reaction has not yet begin (t=0), the concentration of A is 10M.= 1 mol of A[A] = 10 Mt = 0A BAfter 10 seconds, 3 moles of B have formed.[A] = 7 M[B] = 3 Mt = 10= 1 mol of B10 more seconds[A] = 5 M[B] = 5 Mt = 2020 more seconds[A] = 4 M[B] = 6 M40 more seconds[A] = 3 M[B] = 7 Mt = 40t = 80Plotting the data from previous slideReactions Follow a Rate LawThe graph in the previous slide shows that the disappearance of A (formation of B) is not linear.

As [A] decreases, the reaction slows down. This means that reaction rates depend on reactant concentration.

This dependence of rate on concentration suggests that reaction rates follow a rate law, a mathematical expression that ties concentration and rate togetherInstantaneous RatesAlthough the rate of the reaction is constantly changing with reactant concentration, we can determine the instantaneous rate (reaction rate at a specific time and concentration)

Instantaneous rate at t=0 is the initial rate

We can determine the instantaneous rate by taking the slope of the tangent at the point of interest

Note: a tangent line is linear and ONLY touches the point in question. It does NOT cross the curveInstantaneous rate of disappearance of A at t=20 secTangent at t = 20s, [A] = 5MRates and StochiometryIn the previous example (A---->B), we had 1:1 stoichiometry. Thus, at any given time, the rate of disappearance of A equals the rate of formation of B. If the stoichiometry is NOT 1:1, we have a much different situation, as shown below:

As you can see, 2 moles of HI are consumed for every 1 mole of H2 and 1 mole of I2 formed. Thus, the disappearance of HI is twice as fast as the appearance of the products.

Example: N2O5(g) ----> 2NO2(g) + O2(g)

Looking at average ratesaverage rate of disappearance after 10 minutesaverage rate of disappearance after 100 minutes

N2O5(g) ----> 2NO2(g) + O2(g)fastslowRate LawsWe see that reducing reactant concentration lowers the reaction rate, but to what extent? What is the mathematical correlation?

The equation that relates the concentration of the reactants to the rate of reaction is called the rate law of the reaction.

We can derive the rate law of a reaction by seeing HOW THE REACTION RATE CHANGES WITH REACTANT CONCENTRATION.

For any reaction aA + bB ----> cC + dD

In this expression, k is the rate constant, m and n are reaction orders. Reaction Orders and the Method of Initial RatesLets go back to the previous reaction:Below is a table of data, showing the initial reaction rate as a function of the starting concentration of N2O5 (g). We perform multiple experiments to collect enough data to determine our rate law.

We see that when we double [N2O5]o, the rate also doubles. When we quadruple [N2O5]o, the rate quadruples. Thus, the rate is directly proportional to [N2O5]o by the rate constant, k.

This means that the reaction is FIRST ORDER WITH RESPECT TO [N2O5] (m=1). We can write the rate law as:N2O5(g) 2NO2(g) + O2(g)Experiment[N2O5]o (M)Rate, M/s10.010.01820.020.03630.040.072Reaction OrdersThe overall reaction order is the sum of the individual reaction orders. In our previous example, there was only one reactant, so the overall order is 1 (1st order reaction).

We can easily solve for k by plugging in any corresponding rate and concentration. Lets plug in the values from run # 1Run[N2O5]o (M)Rate, M/s10.010.01820.020.03630.040.072Rate Laws/Reaction OrdersReaction orders must be determined experimentally. You can not assume based on the stoichiometry.

When you have multiple reactants, you must determine the reaction order of each one. To do this, you must vary the concentration of only one reactant at a time while holding the others fixed.

Lets attempt to determine the rate law for the reaction below:2NO(g) + O2(g) ---> 2NO2Example: 2NO(g) + O2(g) ---> 2NO2Using the data below, determine the rate law of this reaction in the form: Experiment[NO]o (M)[O2]o (M)Rate (M/s)1.0126.01252.82 x 10-22.0252.02501.13 x 10-13.0252.01255.64 x 10-2This time, we have two reactants. Lets start by determining the value of m. To do so, we hold [O2]o fixed and vary [NO]o. This will show how the rate depends on [NO]o.

In experiments #1 and #3, [O2]o is fixed, so we will use these to find m.Experiment[NO]o (M)[O2]o (M)Rate (M/s)1.0126.01252.82 x 10-22.0252.02501.13 x 10-13.0252.01255.64 x 10-2Remember, rate is proportional to [NO] by the power m. The factor of change in the rate is equal to the factor of change of [NO] to the mth power: factor of rate changefactor of change in [NO]orderm = 1The reaction is 1st order with respect to [NO]Run[NO]o (M)[O2]o (M)Rate (M/s)1.0126.01252.82 x 10-22.0252.02501.13 x 10-13.0252.01255.64 x 10-2factor of rate changefactor of change in [O2]ordern = 1The reaction is 1st order with respect to [O2] and 2nd order overall.Now we can find n by varying [O2]o and holding [NO]o fixed. We can use experiments #2 and #3 for this. This will show how the rate depends on [O2]o.Pay Attention to the Units of k, As They Change with Overall Reaction OrderThe rate constant, k, is the constant of proportionality between rate and concentration.

Higher values of k = faster reactions

It is important to note that the units of k depend on the overall reaction order.

Ex: Rate is always in molarity per unit time (sec, hr, etc). Concentration is always M (mol/L). Thus, we have:

Recall for a 1st order reaction:

Units of k for a 2nd order reactionUnits of k for a 1st order reactionExampleDetermine the relative (m & n) and overall (m+n) reaction order of the reaction below. Then, derive the rate law and determine the value of k (with correct units)Experiment[NO2]o (M)[CO]o (M)Rate (M/s)1.0300.2001 x 1052.0900.2009 x 1053.300.04001 x 1074.300.08001 x 107Tripling [NO2] causes the rate to increase nine-fold. This means that the rate is proportional to the square of [NO2], so the reaction is second order with respect to NO2 (n=2). Doubling [CO] does nothing. Thus, the rate does not depend on [CO], and is zero order with respect to CO (m=0). Overall 2nd order.k = 1.11 x 108 M-1s-1Example: 2NO(g) + Br2(L) ---> 2NOBr (g)Experiment [NO]o[Br2]oRate (M/s)10.100.202420.250.2015030.100.5060Using the information below, determine the rate law of this reaction in the form:1.) Find m. We can use runs 1 & 2:2.) Find n. We can use runs 1 & 3m = 2n = 1The reaction is 2nd order with respect to [NO], 1st order with respect to Br2, and the reaction is overall 3rd order.20Determining the Overall Rate Order of A Reaction GraphicallyAs we have shown, a first-order reaction depends on the concentration of a single reactant to the 1st power. For the reaction: A----> products

Using calculus, we can convert this to:

This equation is in y = mx + b form. Therefore, for any 1st order reaction, the plot of the natural log of [A]t vs time will be linear. The slope of the line will be k.natural log of concentration at time tnatural log of starting concentrationrate constanty axism (slope)btimex axisPlotting 1st Order Reactions

btime values on x-axisslope = -kunits: s-1natural log of [A]t on y-axisDetermining the Overall Rate Order of A Reaction graphicallyA second-order reaction depends on the concentration of [A] to the 2nd power. For the reaction: A ----> B

Therefore, for any 2nd order reaction, the plot of the inverse of [A]t vs time will be linear. The slope of the line will be k.

ymxbPlotting a 2nd Order Reaction

bslope = kunits = M-1 s-1time values on x-axis1/[A]t on y-axis24Determining Overall Rate Order From Plotting Time-Dependent Data

We can determine if a process is first or second order by plotting the data against both equations. Which ever fitting method yields a linear plot gives the overall order. not linear:NOT 1st orderlinear!2nd order