collision model, energy diagrams & arrhenius equation section 7 chemical kinetics chapter 12
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Collision Model, Energy Diagrams & Arrhenius
EquationSection 7
Chemical KineticsChapter 12
aA + bB products Rate = k [A]n [B]m
How does temperature affect rate?
Collision Model
Molecules must collide to react
Increase temp.; increases frequency of collision
Only a small fraction of the collisions produces a reaction.
Why?
Threshold energy = activation energy = the energy that must be over come to produce a chemical reaction
Collision Model
Ea
2BrNO NO + Br2 Rate = k [A]n [B]m
2Br-N bonds must be broken, the energy comes from the kinetic energy of the molecules.
E has no effect on rate – rate depends on the size of the activation energy
Energy Diagram
At any temperature only a fraction of collisions have enough energy to be effective
Energy and Temperature
Possible orientations for a collision between two BrNO molecules
Collision Model
The fraction of effective collisions increases exponentially with temp.
# of collisions Ea = (total collisions) e-Ea/RT
e-Ea/RT = fraction of collisions with E Ea at T
Collision Model
Runiversal gas constant
8.3145 J/K• mol
Rate also depends on molecular orientation
Successful collisions
1. Energy Ea
2. Correct orientation
k = z p e-Ea/RT
z = collision frequency
p = steric factor (always less than one) reflects the fraction of collisions with effective orientation
A = frequency factor, it replaces zp. A=zp
k = A e-Ea/RT
Collision Model
k = A e-Ea/RT
Take the natural log of each side
For a reaction that obeys the Arrhenius equation, the plot of ln (k) vs 1/T give a straight line.
slope = -Ea /R
y-intercept = ln (A)
Most rate constants obey the Arrhenius equation which indicates that the collisions model is reasonable.
Arrhenius Equation
y = m x + b
ln (k) = R
Ea 1
T+ ln (A)x
slopey-intercept
Arrhenius Equation
Take ln(k2) – ln(k1)
Use algebra as on page 557.
ln (k) = R
Ea 1
T+ ln (A)x
ln (k) = R
Ea 1
T+ ln (A)x
11
22
k2 R
Ea 1
T1
_k1
1
T2
ln
=
The values of k1 and k2 measured at T1 and T2 can be used to calculate Ea
The reaction (CH3)3CBr + OH- (CH3)3COH + Br-
in certain solvent is first order with respect to (CH3)3CBr and zero order with respect to OH-. In several experiments, the rate constant k was determined at different temperatures. A plot of ln(k) vs. 1/T was constructed resulting in a straight line with a slope of -1.10 x 104 K and y-intercept of 33.5. Assume k has units of s-1.
a. Determine the activation energy for this reaction.
b. Determine the value of the frequency factor A
c. Calculate the value of k at 25 C.
End of Chapter Exercises #58
here
ANSWER
The activation energy for the decomposition of HI(g) to H2 (g) and I2 g) is 186 kJ/mol. The rate constant at 555 K is 3.52 x 10-7 L/mol•sec. What is the rate constant at 645 K?
End of Chapter Exercises #59
9.5 x 10-5 L/mol•sec
ANSWER
A certain reaction has an activation energy of 54.0 kJ/mol. As the temperature is increased form 22C to a higher temperature , the rate constant increases by a factor of 7.00. Calculate the higher temperature.
End of Chapter Exercises #61
51C
ANSWER
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