Chemical Kinetics - Summary General Chemistry Ii PDF

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Chapter 15 Chemical Kinetics

1. The rate of chemical reaction – the change in the amounts of reagents (or products) divided by the change in time. For the reaction of hydrogen and iodine, H2(g) + I2(g)  2 HI(g) , the reaction rate is defined as 1 [HI] [H 2 ] [I 2 ]   2 t t t Rate= Note that reaction rate is a positive number, therefore, negative sign must be added when reagents are used (because their concentrations decrease) and positive sign when products are used (because concentrations of products increase). Factor ½ is due to stoichiometry because two moles of HI are made from one mole of H2 and I2. Concentration of hydrogen and the reaction rate (magnified 25 times) for the above reaction, is shown below 

2. The instantaneous reaction rate - this is the slope of the concentration curve plotted against the reaction time. For general reaction: aA  bB  cC  dD , the reaction rate is defined as: 1  [A] 1  [B] 1  [C] 1  [D]     b t c t d t Rate = a  t 3. Measuring reaction rates  determine concentration of reagents or products as a function of time, using suitable analytical techniques, for example, absorption of light or chromatography. 4. The rate laws: The reaction rate depends on concentration of reagents. Let’s consider reaction A  products (for example, decomposition of sucrose into glucose and

fructose in acidic solution). We also assume that the rate of incverse reaction is very small. Then the rate law is: n Rate = k [A]

n = 0 - the reaction is zero order n = 1 - the reaction is first order n = 2 - the reaction is second order Examples: (a) zero-order reaction: sublimation is a zero order reaction: A(solid)  A(gas) because sublimation takes place from the surface, so the number of available molecules at a given time is constant. (b) first order reaction – cyclopropane  CH3—CH==CH2 (c) second order reaction: NO2(g)  NO(g) + O(g) 5. Overall reaction order: For reaction aA + bB  cC + dD , the reaction rate is defined as: n m Rate = k [A] [B] , and n + m is the overall reaction order. Examples: NO2(g) + CO(g)

 NO(g) + CO2(g)

H2(g) + I2(g)  2 HI (g)

Rate =

Rate =

k[H2 ]1 [I2 ]1

2H2(g) + 2 NO(g)  N2(g) + 2H2O(g) Rate =

k[NO2 ]2 [CO]0

overall order = 2

overall order = 2 k[H2 ]1 [NO]2

overall order = 3

6. Rate laws and concentration of reagents as a function of time: a) Zero-order reaction

Rate =



[A] k[A]0 t

[A]t  kt [A]t 0

[A] t1/2  t 0 2k

b) First-order reaction  Rate =

[A] k[A]1 t

ln[A]t  kt  ln[A]t0

t1/2 

ln 2 0.693  k k

c) Second order reaction:

Rate =



[A] k[A]2 t

1 1 kt  [A]t [A]t0

t1/2 

1 k [A]t0

7. Graphical representation

8. Determine the reaction rate using MS Excel: graph concentration vs. time. If you get a straight line, than the reaction is zero-order. If not, then the reaction could be first or second order. Then plot logarithm of concentration vs. time. If you get a straight line,

then the reaction is first order. If not, then graph inverse of the concentration vs. time. If you get a straight line with POSITIVE slope, then the reaction is the second order. Example: N2O5(g)  NO2 (g) + NO3(g) Time(s)

0

[N2O5] (M) 1.00 0

25

50

75

100

125

150

175

200

0.82 2

0.67 7

0.55 7

0.45 8

0.37 7

0.31 0

0.25 5

0.21 0

9. The effect of temperature on reaction rate: Rate = k [A]n , and the rate constant, k depends on temperature:

k  Ae



Ea RT

- Arrhenius equation

where A is the frequency collision factor and Ea is the activation energy. The activation energy is the energy barrier that reagents must “climb” in order to transform into products.

10. How to measure the activation energy: E 1 E ln k  ln( A)  a  a RT R T k  Ae  Therefore, we must measure the reaction rate constant as a function of temperature, then graph natural logarithm of the rate constant vs. temperature inverse. Example: T(K)

310

320

330

340

350

k(1/s )

0.0043 4

0.014 0

0.042 1

0.118 0.31 6

1 We graph ln(k) vs. T and get the following graph:

1 ln( k )  We see that -1162 T + 32.055. Ea  E  Therefore, R 1162 and ln( A) = 32.055, or: a 139.8 J/(mol), and the frequency factor A = 8.34x1013 collisions per second. 11. Catalysis – heterogeneous catalysis (takes place on a surface of a solid catalyst) and homogeneous catalysis – a catalyst is in the same phase as reagents. Examples – heterogeneous catalysis: car catalytic converters (Pt, Pd, Rh on ceramic support), synthesis of ammonia (Pt on metal support), hydrogenation of plant oils (nickel on ceramic support), cracking of heavy hydrocarbons (aluminum/chromium oxides on ceramic support). Homogeneous: decomposition of ozone, dehydration of alcohols by acids. 12. Enzymes – synthesis of starch and sugars from CO2 and water (+light), synthesis of biological compounds (for example, antibiotics, alkaloids, and numerous other compounds)....