Things to know……

Rate depends on temperatureTemp is the avg. KE

Order depends on rxn mechanismRate is determined by the slow step

Temp affects k Increase temp 10oC, rate doubles

Rxns occur when collisions have sufficient Ea and correct orientation

Factors that can affect rate:Nature of reactants, surface area,

concentration, temp, catalyst, pressure

Reaction Rates How we measure rates.

Rate LawsHow the rate depends on amounts of reactants.

Integrated Rate LawsHow to calc amount left or time to reach a given amount.

Half-lifeHow long it takes to react 50% of reactants.

Arrhenius EquationHow rate constant changes with T.

MechanismsLink between rate and molecular scale processes.

A plot of concentration vs. time for this reaction yields a curve like this.

The slope of a line tangent to the curve at any point is the instantaneous rate at that time.

C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)

The reaction slows down with time because the concentration of the reactants decreases.

C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)

What if the ratio is not 1:1?

H2(g) + I2(g) 2 HI(g)

• Only 1/2 HI is made for each H2 used.

3. 2 A(g) + B(g) <===> 2 C(g) When the concentration of substance B in the

reaction above is doubled, all other factors being held constant, it is found that the rate of the reaction remains unchanged. The most probable explanation for this observation is that

(A) the order of the reaction with respect to substance B is 1(B) substance B is not involved in any of the steps in the mechanism of the reaction(C) substance B is not involved in the rate-determining step of the mechanism, but is involved in subsequent steps(D) substance B is probably a catalyst, and as such, its effect on the rate of the reaction does not depend on its concentration(E) the reactant with the smallest coefficient in the balanced equation generally has little or no effect on the rate of the reaction

Step 1) N2H2O2 <===> N2HO2¯ + H+ fast equilibrium

Step 2) N2HO2¯ ---> N2O + OH¯ (slow) Step 3) H+ + OH¯ ---> H2O (fast) 4. Nitramide, N2H2O2, decomposes slowly in

aqueous solution. This decomposition is believed to occur according to the reaction mechanism above. The rate law for the decomposition of nitramide that is consistent with this mechanism is given by which of the following?

(A) Rate = k [N2H2O2](B) Rate = k [N2H2O2] [H+](C) Rate = (k [N2H2O2]) / [H+](D) Rate = (k [N2H2O2]) / [N2HO2¯](E) Rate = k [N2H2O2] [OH¯]

A rate law shows the relationship between the reaction rate and the concentrations of reactants. For gas-phase reactants use PA instead of [A].

k is a constant that has a specific value for each reaction.

The value of k is determined experimentally.

“Constant” is relative here- k is unique for each rxnk changes with T (section 14.5)

When ln P is plotted as a function of time, a straight line results.The process is first-order.k is the negative slope: 5.1 10-5 s-1.

Graphing ln [NO2] vs. t yields:

Time (s) [NO2], M ln [NO2]

0.0 0.01000 -4.610

50.0 0.00787 -4.845

100.0 0.00649 -5.038

200.0 0.00481 -5.337

300.0 0.00380 -5.573

• The plot is not a straight line, so the process is not first-order in [A].

Does not fit:

A graph of 1/[NO2] vs. t gives this plot.

Time (s) [NO2], M 1/[NO2]

0.0 0.01000 100

50.0 0.00787 127

100.0 0.00649 154

200.0 0.00481 208

300.0 0.00380 263

• This is a straight line. Therefore, the process is second-order in [NO2].

A graph of 1/[NO2] vs. t gives this plot.

Time (s) [NO2], M 1/[NO2]

0.0 0.01000 100

50.0 0.00787 127

100.0 0.00649 154

200.0 0.00481 208

300.0 0.00380 263

• This is a straight line. Therefore, the process is second-order in [NO2].

10. The graph above shows the results of a study of the reaction of X with a large excess of Y to yield Z. The concentrations of X and Y were measured over a period of time. According to the results, which of the following can be concluded about the rate of law for the reaction under the conditions studied? A) It is zero order in [X].B) It is first order in [X].C) It is second order in [X].D) It is the first order in [Y].E) The overall order of the reaction is 2.

 Experiment Initial [NO]

(mol L¯1

Initial [O2]

(mol L¯1

Initial Rate ofFormation of

NO2

(mol L¯1 s¯1) 1 0.10 0.10 2.5 x 10¯4 2 0.20 0.10 5.0 x 10¯4 3 0.20 0.40 8.0 x 10¯3

The initial-rate data in the table above were obtained for the reaction represented below. What is the experimental rate la for the reaction? (A) rate = k[NO] [O2](B) rate = k[NO] [O2]

2

(C) rate = k[NO]2 [O2](D) rate = k[NO]2 [O2]

2

(E) rate = k[NO] / [O2]

Half-life is defined as the time required for one-half of a reactant to react.

Because [A] at t1/2 is one-half of the original [A],

[A]t = 0.5 [A]0.

For a first-order process, set [A]t=0.5 [A]0 in integrated rate equation:

NOTE: For a first-order process, the half-life does not depend on [A]0.

For a second-order process, set [A]t=0.5 [A]0 in 2nd order equation.

First order Second order Second order

Rate Laws

Integrated Rate Laws complicated

Half-life complicated

KE converts to PE during collisions to break bonds.

The transition state (aka activated complex) is the unstable intermediate that forms at the peak of the PE diagram.

Increase Ae, k decreases, and therefore rate decreases.

When temp doubles, all particles speed up (way more than double); therefore, the relationship is not linear, but rather exponential.

When to use it? When given k and time or asked to solve for Ea.

Plot ln k vs 1/T = linear graph Slope of the line = -Ea/R\ Therefore, Ea = -R x slope R = 8.31 J/K mol