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Chemical Kinetics

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Chemical Kinetics

Chapter 12

- Chemical Kinetics
- The study of the rates at which reactant and product concentrations change during a chemical reaction.

- Reaction rates
- Rates at which reactants are consumed and products are formed.

- Consider the following reaction:
N2(g) + O2 2NO(g)

Rate of NO formation:

= =

Rate of N2 consumption:

=

Rate of O2 consumption:

=

Units are expressed as concentration per unit time…

Molarity per second (M/s)

- Consider the following reaction:
N2(g) + O2 2NO(g)

- However, we must account for mole ratios in the balanced equation. From the mole ratios and the previous equations, we deduce that the rate of formation of NO is twice that of the rate of consumption of either O2 or N2. (for every one mole of either, there are two moles of NO formed)
- We represent this ratio by inserting the mole ratios into the denominator of the rate expression.

- Consider the following reaction:
N2(g) + O2 2NO(g)

- SO, the complete kinetics reaction is:
- -2=-2= or:
- -=-= I prefer this expression

- Practice Question:
- Consider the reaction:
N2(g) + H2(g) NH3(g)

Balance the equation and express the complete kinetics reaction equation

- Practice Question:
- Consider the reaction:
N2(g) + 3H2(g)2NH3(g)

Balance the equation and express the complete kinetics reaction equation

-=-=

- Practice Question:
- Consider the reaction:
N2(g) + 3H2(g)2NH3(g)

Balance the equation and express the complete kinetics reaction equation

- - =-= or:
- -2 =- =

- Practice Question:
- Consider the reaction:
N2(g) + 3H2(g)2NH3(g)

Based on your kinetics expression, if under a given set of conditions, the rate of ammonia formation was found to be 0.472 M/s, calculate the rate of consumption of Nitrogen and Hydrogen.

- Based on your kinetics expression, if under a given set of conditions, the rate of ammonia formation was found to be 0.472 M/s, calculate the rate of consumption of Nitrogen and Hydrogen.
- - =- =
- - =
- - = =

- Based on your kinetics expression, if under a given set of conditions, the rate of ammonia formation was found to be 0.472 M/s, calculate the rate of consumption of Nitrogen and Hydrogen.
- - =- =
- - = .236 M/s
- - = = .708 M/s
Even though the ACTUAL rate is negative, we express it as a positive integer because the term consumption means we are using the reactant and the change in concentration will obviously be negative. If the question asks you for concentration change based on time, the answer will be negative.

- Practice problem #2
- Consider the following reaction:
NO(g) + O2 NO2(g)

If the rate of O2 consumption is .033 M/s, what s the rate of formation of NO2 and rate of consumption of N2?

NO2=?

N2=?

- Consider the following reaction:

- Practice problem #2
- Consider the following reaction:
NO(g) + O2 NO2(g)

If the rate of O2 consumption is .033 M/s, what s the rate of formation of NO2 and rate of consumption of N2?

NO2=0.066 M/s

NO =.0165 M/s

- Consider the following reaction:

- Read. Period. Chapter 12, sections 1-5.

- How does concentration affect the rate of reactions?
- How do reactions occur? Think of elastic collisions and Kinetic energy….
- What about the addition of a catalyst?

- Rate law-an equation that defines the relation between reactant concentrations and reaction rate.
- Rate constant- (k) the proportionality constant that relates the rate of reaction to the concentration of reactants.
- Reaction order-a parameter derived from experiments that tells us how reaction rates depends on reactant concentrations.

- We determine the reaction order values by comparing differences in reaction rates to differences in reactant concentrations
- We are not able to do this without concentration values from multiple experiments.
- Consider the reaction:
NO(g) + O2 NO2(g)

- Consider the reaction:
NO(g) + O2 NO2(g)

A + B C

- First, we start out with writing our rate law equation:
Rate = k[NO][O2]---k=[A][B]

- Do not confuse this with writing the equation for calculating reaction rate.
- Do not take into consideration the mole coefficients.
- k = rate constant.

- Second, insert reaction order exponents for the reactants:
- Rate = k[NO]m[O2]n------k[A]m[B]n
- m= reaction order with respect to NO
- n= reaction order with respect to O2

- Next, use the following equations to solve for n or m:
- = )n whereas: log() = n log ()
- n=
- Subscripts 1,2 refer to experiment number, [B] is concentration of arbitrary reactant, could just as easily been [A] or [C]. n is reaction order with respect to reactant B.
- The overall reaction order is the sum of exponents (n + m + …..)

- Given the following experimental data for NO(g) + O2 NO2(g) at 25°C, determine the rate order for NO and O2and the overall reaction order.
- When determining rate order for a reactant, use the experiments in which the other reactant concentrations stay constant.

- Once you solve for n and m, insert exponents into rate law equation to determine overall and partial rate orders.
- For rate order with respect to NO
- m=

- Rate = k [NO]?[O2]?

- Example #2: consider the following reaction:
- 2ClO2(g) + 2OH-(aq) ClO3-(aq) + ClO2-(aq) + H2O(l)
- Given the following experimental data, calculate the overall reaction order, write the rate law expression and calculate k.

- 2ClO2(g) + 2OH-(aq) ClO3-(aq) + ClO2-(aq) + H2O(l)

- Rate order with respect to both reactants is first order. Overall reaction order is second order.
- k=6.32 M/s

- Remember that the rate constant, k, does not change with concentration or pressure…only with a change in temperature or the presence of a catalyst.

- The mathematical connection between temperature, the rate constant (k) and its activation energy is given by the Arrhenius equation:
- k= Ae-Ea/RT

- k= Ae-Ea/RT
- Ea=activation energy
- R = 8.314 J/molK
- T= temperature
- A= frequency factor-the product of collision frequency.
- The function ex so defined is called the exponential function, and its inverse is the natural logarithm, or logarithm to base e.

- Activated complexes represent the midway points in chemical reactions and only exist for a very short period of time.

- The internal energy of an activated complex represents a high-energy transition state of the reaction.
- We can draw an energy profile for a chemical reaction that shows the changes in energy for the reaction as a function of the progress of the reaction.

- If you have not read chapter 12 by now you are behind. Read it, expect a quiz on it sometime this week. And no…I will not help you on it, but you can use your notes.