Reaction Kinetics Flashcards
For the reaction:
A + 2B –> 3P
Where the rate expression is defined as:
- rA = 2 CA^0.5 CB
Write the rate expression for each of the components.
- rB = 4 CB^0.5 CB
rC = 6 CA^0.5 CB
Define ‘conversion’ of a reaction
The ratio between the amount of A reacted and the initial amount of A
Write the general rate expression and derive the expression for a constant volume system
- rA = k CA^a
- rA = dCA/dt
- rA = - 1/V dnA/dt
- rA = nA0/V dXA/dt
OR - rA = CA0 dXA/dt
What is the reaction rate dependent upon?
- Concentration of reactants
- Rate constant, k
- Temperature
- Catalyst
What affects the rate constant, k?
- Temperature
2. Catalyst
Write the Arrhenius equation and write the equation if this was plotted on a graph
k = A e^(-E/RT)
ln k = ln A - (E/R)*(1/T)
Define each of the components in the Arrhenius equation
E = activation energy, the minimum kinetic energy that reactants must have in order to form products
exp = exponential factor, the fraction of collusions that have enough kinetic energy to lead that reaction
A = pre-exponential factor or frequency factor, a measure of the rate at which collisions occurred, irrespective of their energy
k = rate constant, the rate of successful collisions
Write the Arrhenius equation for comparing 2 different temperatures
ln k2/k1 = E/R (1/T1 - 1/T2)
True or false; reactions with a lower activation energy are more sensitive temperature
False
Higher activation energies are more sensitive to temperature changes. This is shown by a steeper gradient on the Arrhenius plot.
Define ‘reaction order’
The index, or exponent, to which its concentration term in the rate equation is raised.
True or false; the stoichiometric coefficients of a reactant influence the order of the reaction
False
There is no relationship between these two factors
In what case will a temperature increase cause the rate of reaction to decrease?
When the reactant degrades at a certain temperature, the rate of reaction will decrease above this temperature
Show how to calculate and compare the fraction of molecules able to react at 2 different temperatures
k1 = A e^(-E/RT1) k1 = \_\_\_\_ A
k2 = A e^(-E/RT2) k2 = \_\_\_\_ A
k2/k1 = Fraction
Show how to calculate and compare the fraction of molecules able to react at when a catalyst is present at a given temperature
k1 = A e^(-E/RT) k1 = \_\_\_\_ A
k cat = A e^(-E/RT)
k cat = ____ A
k cat/k1 = Fraction
Write the general rate equation for a first order, homogenous, gas phase reaction
r = k pA
Show how the rate equation for a homogenous, first-order reaction can be written in terms of:
- Catalyst weight
- Surface of catalyst
- Volume of fluid
- Volume of reactor
- Volume of solid
- rA = 1/W dNA/dt
- rA = 1/S dNA/dt
- rA = 1/V dNA/dt
- rA = 1/Vr dNA/dt
- rA = 1/Vs dNA/dt
Define the difference between the units for rate of reaction for a homogenous reaction vs. heterogeneous
Homogenous = mol/ L min Heterogeneous = mol/ m^2 min
What is an elementary reaction?
A reaction where the orders of the reactants match the stoichiometry of the reaction
True or false: A rate equation that has reaction orders corresponding to the reaction stoichiometry are always elementary reactions
False
It can just be a coincidence
Write the rate equations and the equivalent rate constants for each of the components of the following elementary reaction:
B + 2D –> 3T
- rB = kB CB CD^2
- rD = kD CB CD^2
rT = kT CB CD^2
rB = 1/2 rD = 1/3 rT kB = 1/2 kD = 1/3 kT
Use the initial rates method to determine the rate law using the following data:
A + 2B –> 3C where r = k CA^a CB^b
1 - r = 2.73 - CA = 0.1 - CB = 0.1
2 - r = 6.14 - CA = 0.15 - CB = 0.1
3 - r = 2.73 - CA = 0.1 - CB = 0.2
ln r = a ln CA + b ln CB + ln k
- ln 2.73 = a ln 0.1 + b ln 0.1 + ln k
- ln 6.14 = a ln 0.15 + b ln 0.1 + ln k
- ln 2.73 = a ln 0.1 + b ln 0.2 + ln k
1 - 2 = ln 2.73/6.14 = a ln 0.1/0.15 –> a = 2
1 - 3 = ln 2.73/2.73 = b ln 0.1/0.2 –> b = 0
Sub in to any of 1, 2 or 3
k = 273 L/mol s
r = 273 CA^2 CB^0
Use the isolation method to determine the rate law using the following data:
A + 2B –> 3C where r = k CA^a CB^b
1 - r = 2.73 - CA = 0.1 - CB = 0.1
2 - r = 6.14 - CA = 0.15 - CB = 0.1
3 - r = 2.73 - CA = 0.1 - CB = 0.2
r1 = k CA1^a CB1^b r2 = k CA2^a CB2^b r3 = k CA3^a CB3^b
r1/r2 = (k CA1^a CB1^b)/(k CA2^a CB2^b)
2.73/6.14 = (0.1/0.15)^a
log 2/3 (2.73/6.14) = a
a = 2
r1/r3 = (k CA1^a CB1^b)/(k CA3^a CB3^b)
2.73/62.73 = (0.1/0.2)^b
log 0.5 (2.73/2.73) = b
b = 0
Derive the two integrated rate laws for a first order reaction
ln (CA/CA0) = - kt
e^(ln (CA/CA0)) = e^(- kt)
CA/CA0 = e^(- kt)
CA = CA0 e^(-kt)
How do you determine the rate constant graphically for a first order reaction (hint: using the integrated rate law)
ln (CA/CA0) = - k t
plot t (x-axis) vs. ln (CA/CA0) (y-axis)
gradient = -k
Derive the two integrated rate laws for a second order reaction
1/CA - 1/CA0 = kt
CA = CA0/(1 + k t CA0)
What is half life?
The time taken for the concentration of the reactant to reduce to half of that concentration
Derive the equation for the half life of a first order reaction
ln (CA/CA0) = - k t
t 1/2 = 1/k ln (0.5 CA0/CA0)
t 1/2 = 1/k ln (0.5)
t 1/2 = ln 2/k
True or false: For a first order reaction, the half life is independent of initial concentration
True
Derive the equation for the half life of a second order reaction
1/CA - 1/CA0 = kt
t 1/2 = 1/kCA0
True or false: For a second order reaction, the half life is independent of initial concentration
False
Write the equation for the number of half lives
No. half lives = total time elapsed/half-lives
Write the equation for the fraction of molecules remaining after a given number of half lives
(Amount remaining/Initial amount) = (0.5)^(No. half lives)
Amount remaining/Initial amount) = (0.5)^(Total time elapsed/Half-lives
Outline the method for calculating the integrated rate law for bimolecular reactions
A + B –> P
- rA = k CA CB = - dCA/dt = - dCB/dt
Let the amount of A or B reacted = m = CA0 - CA = CB0 - CB
Therefore:
- rA = dm/dt = k (CA0 - m) (CB0 - m) = - dCA/dt
dm/dt = k (CA0 - m) (CB0 - m)
k dt = 1/(CA0 - m) (CB0 - m) dm
Initial conditon:
m = 0 when t = 0
Use partial fraction integration to determine the integral of the RHS:
1/(AD - BC) ln ((Ax + B)/(Cx + D))
1/(CA0 - CB0) ln ((CB0 CA)/(CB CA0))
Therefore:
1/(CA0 - CB0) ln (CB0CA/CBCA0) = kt
n (CB0CA/CBCA0) = (CA0 - CB0)kt
Derive the integrated rate law for a reversible reaction
A B
CA = [(k’+ke^(k+k’)t)/(k+k’)] CA0
Write the rate equations for each of the components in the following reversible and concurrent reactions:
A B
A —> C
- rA = k1 CA + k2 CA
rA = k-1 CB
rA = k-1 CB - (k1 + k2) CA - rB = k-1 CB
rB = k1 CA
rB = k1 CA - k-1 CB
rC = k2 CA
Write the equation for the equilibrium constant for a reversible reaction
A B
K = CB,eq/CA,eq K = [kCA0/(k+k')] / [k'/(k+k')] CA0 K = k/k'
What happens to the integrated rate law for a reversible reaction as t –> infinity?
A B
Concentrations –> Equilibrium
Hence: e^(k+k’)t –> 0
CA,eq = [(k'+ke^(k+k')t)/(k+k')] CA0 CA,eq = [k'/(k+k')] CA0
CB,eq = CA0 - CA,eq CB,eq = kCA0/(k+k')
