Integrated Rate Laws Finally a use for calculus! Text 692019 and Questions to 376071.

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Integrated Rate Laws Finally a use for calculus! Text 692019 and Questions to 376071

What is a rate? Text 692019 and Questions to 376072

A rate as a derivative Text 692019 and Questions to 376073

Lets look at the rate law Text 692019 and Questions to 376074

Solving the equation Text 692019 and Questions to 376075

Solving the equation Text 692019 and Questions to 376076

Solving the equation Text 692019 and Questions to 376077

What it means… Text 692019 and Questions to 376078

If you know k and the initial concentration, you could calculate the concentration at any time. For example, if I know k=0.015 s -1 and I start with 0.250 M A, how much A is left after 1 minute? Beware the units. 1 minutes = 60 seconds. Since k is in s - 1, I need my time to be in seconds. Plug and chug, baby! Text 692019 and Questions to 376079

Using the integrated rate law Text 692019 and Questions to 3760710

Compare the integrated rate law to the rate law Text 692019 and Questions to 3760711

Other uses of the integrated rate law Text 692019 and Questions to 3760712

For example, suppose I monitor [A] Time (seconds)[A] (M) 0 0.25 10 0.20 20 0.17 60 0.075 Since this is a first order reaction, the data should obey my integrated rate law. So I plot the ln[A] vs time and I should get a straight line. Text 692019 and Questions to 3760713

For example, suppose I monitor [A] Time (seconds)[A] (M)ln[A] 0 0.25-1.386 10 0.20-1.609 20 0.17-1.772 60 0.075-2.590 Now, I plot the last column against the first column and put the best fit straight line on it. Text 692019 and Questions to 3760714

LOOK! ITS A POINT! ITS A PLANE! NO!!! ITS A STRAIGHT LINE! Text 692019 and Questions to 3760715

So, whats the rate constant? y = -0.02x - 1.3863 ln[A] final = - kt + ln[A] t=0 m= slope=-0.02 m=-k k=-(-0.02)=0.02 s -1 So, if I KNOW its a 1 st order reaction, I can make a graph to find the rate constant. I can also make a graph to find out IF it is 1 st order. Text 692019 and Questions to 3760716

Different reaction 2 H 2 + O 2 2 H 2 O Time (seconds)[H 2 ] (M)ln[H 2 ] 0 0.500-0.69315 10 0.300-1.20397 20 0.200-1.60944 60 0.100-2.30259 Now, I plot the last column against the first column and put the best fit straight line on it to see IF IF IF it is actually a straight line. Text 692019 and Questions to 3760717

NOT a straight line – NOT a 1 st order reaction! Text 692019 and Questions to 3760718

Is it or isnt it a straight line? A.It is a straight line B.It is NOT a straight line C.I cant tell without error bars D.I really dont care its Monday E.Your mother! Text 692019 and Questions to 3760719

This works for other orders of reaction also. Text 692019 and Questions to 3760720

Hey! Its 2 nd order! Text 692019 and Questions to 3760721

Also, theres the rare zeroth order Text 692019 and Questions to 3760722

Those are the easy ones For more complicated mixed orders like: Rate = k[A][B] The math gets much more complicated, so well ignore them until you become a chemistry major. But you can do a similar thing. Text 692019 and Questions to 3760723

But a lot of reactions fall into those three categories. Text 692019 and Questions to 3760724

How do we use this? Time[N 2 (g)] (M) 0 min0.40 5 min0.25 10 min0.17 30 min0.04 60 min0.005 N 2 (g)+ 3 Cl 2 (g) 3 NCl 3 (g) Given the following data, determine the rate law. Text 692019 and Questions to 3760725

GRAPH IT! Text 692019 and Questions to 3760726

Graph It! Time[N 2 (g)] (M)Ln([N 2 ])1/[N 2 ] 0 min0.40-0.9162.5 5 min0.25-1.3864.0 10 min0.17-1.7725.88 30 min0.04-3.21925 60 min0.005-5.298200 N 2 (g)+ 3 Cl 2 (g) 3 NCl 3 (g) Given the following data, determine the rate law. Text 692019 and Questions to 3760727

Try all 3 and see which one fits! Text 692019 and Questions to 3760728

Not 0th order – unless it was a sloppy experiment Text 692019 and Questions to 3760729

Maybe 2 nd order – its not a horrible fit. Text 692019 and Questions to 3760730

Hey, Goldilocks! It Fits 1 st order! Text 692019 and Questions to 3760731

What if I dont want to or cant make a graph? A.Find someone who can make a graph. B.Copy the answer from the person next to me. C.Calculate the rate of the reaction and see if the rate is constant or if the ln(rate) is constant or 1/rate is constant. D.Calculate the slope between data points and see if they are constant. Text 692019 and Questions to 3760732

What if I dont want to make a graph? Time[N 2 (g)] (M) 0 min0.40 5 min0.25 10 min0.17 30 min0.04 60 min0.005 N 2 (g)+ 3 Cl 2 (g) 3 NCl 3 (g) Given the following data, determine the rate law. Text 692019 and Questions to 3760733

3 possibilities Text 692019 and Questions to 3760734

k is the rate CONSTANT and its the slope of the line Text 692019 and Questions to 3760735

Slope is all over the place except 1 st order Time (min) [N 2 (g)] (M) slope– 0 th orderslope - 1 st orderK – 2 nd order 00.40 50.25 0.0160.0770.376 100.17 0.00650.07230.96 300.04 0.001170.0695.83 600.005 N 2 (g)+ 3 Cl 2 (g) 3 NCl 3 (g) Given the following date, determine the rate law. Text 692019 and Questions to 3760736

Problem recognition Whats the tell? How do I know how to handle the problem? Text 692019 and Questions to 3760737

Method of initial rates – Rates measured for different initial mixes The reaction: 2 I - (aq) + S 2 O 8 2- (aq) 6 I 2 (aq) + 2 SO 4 2- (aq) was studied at 25° C. The following results were obtained for the rate of disappearance of S 2 O 8 2- [I-] 0 (M) [S 2 O 8 2- ] 0 (M)Initial rate (M/s) 0.0800.04012.5x10-6 0.0400.0406.25x10-6 0.0800.0206.25x10-6 0.0320.0405.00x10-6 0.0600.0307.00x10-6 Text 692019 and Questions to 3760738

Integrated rate law – concentration at different times Time[N 2 (g)] (M) 0 min0.40 5 min0.25 10 min0.17 30 min0.04 60 min0.005 N 2 (g)+ 3 Cl 2 (g) 3 NCl 3 (g) Given the following date, determine the rate law. Text 692019 and Questions to 3760739

Does that make sense? A.Yes B.No C.Maybe Text 692019 and Questions to 3760740

Once I know the order, hows it work…? Once I know the order of the reaction, I can use the integrated rate law to determine the concentration at any time. Text 692019 and Questions to 3760741

The following reaction is 1 st order in Cl 2 and 1 st order overall. H 2 (g) + Cl 2 (g) 2 HCl(g) 2 M H 2 and 2 M Cl 2 was placed in a 5 L flask at 298 K. The initial rate was 3.82x10 -3 M/s. What was the rate after 10 minutes? How much HCl had been made after 10 minutes? Text 692019 and Questions to 3760742

Do I know the rate constant? A.Yes B.No C.Not directly but implicitly D.I have no clue E.You look beautiful today Text 692019 and Questions to 3760743

As soon as Im talking about TIME, its an integrated rate law problem. The order of the reaction was given. This actually tells me two things: The Rate Law The Integrated Rate Law Text 692019 and Questions to 3760744

Text 692019 and Questions to 3760745

Text 692019 and Questions to 3760746

The following reaction is 1 st order in Cl 2 and 1 st order overall. H 2 (g) + Cl 2 (g) 2 HCl(g) 2 M H 2 and 2 M Cl 2 was placed in a 5 L flask at 298 K. The initial rate was 3.82x10 -3 M/s. What was the rate after 10 minutes? How much HCl had been made after 10 minutes? Text 692019 and Questions to 3760747

Text 692019 and Questions to 3760748

Text 692019 and Questions to 3760749

Text 692019 and Questions to 3760750

A.Yes B.No C.Maybe D.You look like crap E.You look beautiful Text 692019 and Questions to 3760751

The following reaction is 1 st order in Cl 2 and 1 st order overall. H 2 (g) + Cl 2 (g) 2 HCl(g) 2 M H 2 and 2 M Cl 2 was placed in a 5 L flask at 298 K. The initial rate was 3.82x10 -3 M/s. What was the rate after 10 minutes? How much HCl had been made after 10 minutes? Text 692019 and Questions to 3760752

In a word… A.Exploitation B.Death C.Life D.Stoichiometry E.Integration Text 692019 and Questions to 3760753

Rate = k[Cl 2 ] Rate = 1.92x10 -3 s -1 (0.632 M) = 1.2135x10 -3 M/s Text 692019 and Questions to 3760754

How much HCl? Text 692019 and Questions to 3760755

Cl 2 (g) + H 2 (g) = 2 HCl (g) Text 692019 and Questions to 3760756

Another fun little rate thing… Half-life! For a reaction, you start with a lot of reactants and you end up with less reactants and more product. The amount of reactants should always be decreasing. Lets look at our earlier example… Text 692019 and Questions to 3760757

Not 0th order – unless it was a sloppy experiment Text 692019 and Questions to 3760758

Hey, Goldilocks! It Fits 1 st order! Text 692019 and Questions to 3760759

So, it is 1 st order. It must obey the first order rate equation. The concentration of the reactants should be asymptotically approaching zero. So if I start with the maximum A, soon I have 90% left, then 80% left, then 70% left…eventually 50% left. The time it takes for ½ (50%) of the A to react is called the half-life. Text 692019 and Questions to 3760760

Text 692019 and Questions to 3760761

Text 692019 and Questions to 3760762

t 1/2 = 2 hours So, lets say I start with 1 mol of Cl 2. In 2 hours, how much Cl 2 is left? A.1 mol B.0.75 mol C.0.50 mol D.0.25 mol E.I dont know enough to calculate it. Text 692019 and Questions to 3760763

t 1/2 = 2 hours Suppose I come back the next morning and find that there is only 0.016 mol Cl 2 left. In 2 hours, how much Cl 2 is left? A.0.016 mol B.0.008 mol C.0.004 mol D.It depends on how much the rate has slowed down as the Cl 2 decreased. E.You look FAB-ulous! Text 692019 and Questions to 3760764

1 st order is special… Radioactive decays show 1 st order kinetics. Thats why you hear half-life when people are talking about reactivity. But half-life actually applies to any reaction: its the time it takes for ½ the reactants to react! Text 692019 and Questions to 3760765

It also doesnt have to be ½ life. Suppose Im arrogant, obstinate, and just a general pain in the patootie… I insist on using t 9/10 – the time it takes for 90% of the reactants to react. Again, if it is first order…. Text 692019 and Questions to 3760766

Text 692019 and Questions to 3760767

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