1. Chem 1310: Introduction to physical chemistry Part 2c: integrated rate laws

2. The integrated rate law If we know how the rate depends on all concentrations, we can predict how it will change over time. This is called the "integrated rate law" At each point we know concentrations, hence the rate, hence we can predict concentrations at the next moment: rate =  [Z]/  t  [Z] = rate*  t new [Z] = old [Z] +  [Z] = old [Z] + rate*  t

3. The integrated rate law A rigorous derivation of integrated rate laws requires calculus, but we can understand and use the results without using calculus. We will discuss only three cases: First-order (most important by far!) Second-order Zero-order

4. The integrated rate law for a first-order reaction For a first-order rate law rate = k [A] the integrated rate law is [A] t = [A] 0 e -k t ("exponential decay") or equivalently ln [A] t = ln [A] 0 - k t So, plotting ln [A] t vs t should give a straight line with slope -k and intercept ln [A] 0.

5. Checking for the crystal violet reaction [CV + ] vs tln [CV + ] vs t ln plot is nicely linear...

6. First-order integrated rate laws First-order rate laws are very common, not only in chemistry. They express a situation where the chance of something happening to each of a set of objects is constant in time and the same for each object, independent of what happens to the other objects.

7. Exponential decay rate law Reasonable examples: People not winning the lottery, if they play consistently for the same amount of money and the lottery company doesn't "change the rules". People never losing at blackjack, provided they and the dealer keep the same strategy (and there are infinitely many cards). Decay of radioactive nuclei. Houses not being hit by a meteorite (assuming all houses have the same roof area etc).

8. Exponential decay rate law Keep in mind: Molecules are perfect at playing the game of chance. They never react "all at the same time". Each molecule has at any time a certain probability to undergo the reaction. If the probability is high, most molecules will react quickly.

9. Exponential decay and half-life A simple exponential-decay curve.

10. Exponential decay and half-life At time t ½, the concentration is half of its original value.

11. Exponential decay and half-life We take the remainder of the curve (past t ½ )...

12. Exponential decay and half-life Expand it vertically by a factor of 2...

13. Exponential decay and half-life Perfect fit! So every part of the curve has the same "shape", and every time we wait a time of t ½ the concentration halves. For other decays, the half-life is not constant. And move it back to t = 0...

14. Relation between rate constant and half-life

15. Using exponential decay (ex 43, p 661) SO 2 Cl 2 decomposes into SO 2 and Cl 2 in a first-order reaction: SO 2 Cl 2  SO 2 + Cl 2 The half-life at 600K is 1.47*10 4 s (4.08 hr). If you begin with 1.6*10 -3 mol in a 2-L flask, how long will it take till only 1.2*10 -3 mol is left? And if we do the same experiment in a 1-L flask? How much is left after 5 hr?

16. Decomposition of SO 2 Cl 2 The rate constant is k = 0.693/t ½ = 0.0000472 s -1 and we have Reducing the amount from 1.6*10 -3 mol to 1.2*10 -3 mol reduces the concentration from 0.8*10 -3 mol/L to 0.6*10 -3 mol/L, so we want [A] t /[A] 0 = 0.6/0.8 = 0.75.

17. Decomposition of SO 2 Cl 2 We now calculate t from: If we do the reaction in a smaller flask, begin and end concentrations will increase (*2), but their ratio stays the same  same answer.

18. Decomposition of SO 2 Cl 2 After 5 hr (t = 18000 s): [A] t = [A] 0 e -0.000047*18000 = [A] 0 *0.429 = 0.000343 mol/L Amount remaining: [A] t * 2 L = 0.69*10 -3 mol.

19. For a second-order rate law rate = k [A] 2 the integrated rate law is So, plotting 1/[A] t vs t should give a straight line with slope k and intercept 1/[A] 0. The integrated rate law for a second-order reaction

20. The integrated rate law for a zero-order reaction For a zero-order rate law rate = k (constant!) the integrated rate law is [A] t = [A] 0 - k t (only valid as long as k t <= [A] 0 ). Plot [A] t vs t: slope -k, intercept [A] 0. Zero-order reactions are rare!

21. Using graphs to analyze rate laws t (s)[A] (mol/L) 0.01.000 10.00.833 20.00.714 30.00.625 40.00.556 50.00.500 60.00.455 80.00.385 100.00.333 [A] vs tln[A] vs t 1/[A] vs t Not zero-orderNot first-order Looks like second-order