Kinetics 1: Rate laws, theories concerning rates

Question 1

• How is rate defined?
• Example of unit?
• Express rate as a derivative

Answer 1

• Change in concentration per time
• eg dm^-3 s^-1
• dc/dt

Question 2

Why is rate usually expressed as a derivative?

Answer 2

Rate often varies with time as well as just conc, so defining it over a very small time interval dt removes this variation.

ie on a graph of conc against time, rate becomes the gradient of a tangent at a given conc value

Question 3

Define the rate in terms of A and B:

A –> B

Answer 3

r = -(1/V_A)(d[A]/dt)

r = (1/V_B)(d[B]/dt)

defining in this way means rate is consistent and always +ve. can also define rate in terms of the rate law, ie r = k[A]

Question 4

H₂ + Br₂ –> 2HBr has rate law r = (K_a[H₂][Br₂]^3/2)/([Br₂] + K_b[HBr])

Which orders can’t be defined?

Answer 4

Overall order

Order wrt Br₂

Question 5

Rate constants often vary with temp, many according to the Arhenius equation.

• Give the eq
• Define the components, giving units

Answer 5

K_T = Aexp(-E_a/RT)

• K_T = rate constant, units vary
• A = A-factor / pre-exponential factor* same units & dimensions as K, since exp term is dimensionless
• E_a = act energy in kJ mol^-1
• R = gas constant, in kJ [temp unit]^-1 mol^-1 (so that overall, exp term is dimensionless)
• T = temp, K

* A is the collision rate per unit reactant concentration. A is also the rate constant at infinite temp, since 1/T = 0 when lnK = lnA. A is independent of or weakly dependent on temp

Question 6

Range of typical values for E_a

Answer 6

10-200 kJ mol^-1

Question 7

Value of RT at room temp

Answer 7

~2.4 kJ mol^-1

Question 8

Derive the straight-line Arhenius equation rearrangement

Answer 8

K = Aexp(-E_a/RT)

lnK = ln(Aexp(-E_a/RT) = lnA - E_a/RT = -E_a/R x 1/T + lnA

Question 9

How is K_eq defined?

Answer 9

Rate constants of the forward reaction / that of the backward reaction

derives from the fact that, by definition of eq, K_forward x bla = K_backward x bla

Question 10

What is a potential energy surface? Where on the surface to stable molecules sit?

Answer 10

A plot of potential energy as a function of the positions of all atoms in a reaction system.

Stable molecules are in potential energy minima.

Multi-dimensional apart from simple molecules

Question 11

The reaction pathway involving the smallest energy expenditure is favoured.

• What is this pathway called?
• What is the potential energy maximum on this pathway called?
• What fraction of molecules have at least this energy, ie the activation energy?
• Why can we assume all reactants react via the minimum energy route?

Answer 11

• The reaction coordinate (i.e. that 2D graph of energy against reaction progress)
• Transition state
• exp(-E_a/RT) according to Boltzmann distribution
• The fraction is an exponential function of energy, so increasing E by a small amount (as would be the caase in choosing a different pathway) would give a negligable fraction of molecules with the required energy

Question 12

Define transition state.

Answer 12

The point of highest energy on the pathway from reactants to products, in which there are partially broken and partially formed bonds.

Question 13

For a reaction with E_a = 50 kJ mol^-1, what fraction of molecules have sufficient energy to attain the transition state?

Answer 13

Fraction = exp(-E_a/RT)

Recall RT = 2.4 kJ mol-1 at room temp

so fraction = exp(-50/2.4) ~ 1 in 10⁹

Question 14

Gas kinetic theory is based on Newtonian mechanics. What 4 assumptions does it make?

What do the assumptions mean for the distribution of particle energies and speeds in a sample?

Answer 14

Assumptions

• Reactions occur whenever molecules collide, since particles are spherical and orientation is irrelevant (biggest defect of this theory)
• Size of particles << distance between particles
• KE causes random movement of particles
• All collisions are elastic (energy conserved)

Implication

Energies and speeds of particles in a macroscopic sample are in a Boltmann/Maxwell distribution.

Question 15

Gas kinetic theory

• Expression for mean speed, give unit
• Bimolecular collisions depend on the relative motion of the molecules. Expression for mean relative speed?

Answer 15

ĉ = √(8K_BT/πm) in ms^-1

K_B is Boltzmann constant, m = particle mass in kg

ĉ = √(8K_BT/πμ) in ms^-1

Where reduced mass, μ = (m_Am_B/m_A+m_B)

μ in AMU, atomic mass units

Question 16

Collision rate, Z_AB, between molecules A and B, represented by spheres of radii r_A and r_B:

Z_AB = C_AC_Bσĉ_rel

• Give the expression for σ and define it
• Given that units are all SI, derive units of Z_AB

Answer 16

• Collision cross section, σ = π(r_A + r_B)² (σ is the area around A for which B is close enough to react with it)

Z_AB = C_AC_Bσĉ_rel = (C_AC_B)(π(r_A + r_B)²)(√(8k_BT/μ)

Radii in m, μ in kg, speeds in molecules m^-3

So units = collisions m^-3 s^-1

Question 17

Derive an expression for the 2nd order rate constant, assuming a bimolecular reaction: A + B –> C.

Answer 17

Get the first expression for r

Fraction of collisions with at least Ea = exp(-Ea/RT)

So #successful collisions per volume per time = Z_ABexp(-Ea/RT)

Each collision gives one product molecule, so #mol product per volume per time = LZ_ABexp(-Ea/RT) where L = avo number

So r = (1/L)Z_ABexp(-Ea/RT)

= (1/L)(C_AC_Aσĉ_rel)

Get the 2nd expression for r

Assume 1st order in A and B, so r = k(C_A x C_B)/L here have divided by avo number so that concs are in moles per vol, so rate is in mol per vol per time

Combine expressions

(1/L)(C_AC_Aσĉ_rel) = k(C_A x C_B)/L

Rearrange –> k = Lσĉ_relexp(-Ea/RT) = Lσ(√(8k_BT/μ))​exp(-Ea/RT) in m³ mol^-1 s^-1

Question 18

bimolecular reaction: A + B –> C.

Compare the following expression for the 2nd order rate constant k to the Arhenius equation, in order to derive an expression for the A-factor.

k = Lσĉ_relexp(-Ea/RT)

Answer 18

k = L(π(r_A + r_B)²)(√(8k_BT/μ))​exp(-Ea/RT)

Arhenius k = Aexp(-Ea/RT)

So A = L(π(r_A + r_B)²)(√(8k_BT/μ))

this is why it’s defined as collision rate per unit reactant concentration

Question 19

A = L(π(r_A + r_B)²)(√(8k_BT/μ))

This equation predicts A to be temp-dependent. Why does this contribute negligably to the expression for r?

Answer 19

strong temp dependence of the exp factor overpowers that of A

Question 20

A = L(π(r_A + r_B)²)(√(8k_BT/μ))

This expression can be tested experimentally. Does collision theory generally underestimate or overestimate A, and why?

Answer 20

Overestimates it, because collision theory assumes all collisions with Ea give successful reaction, whereas in reality orientational effects mean the number of successful collisions is lower.

Question 21

Define and give the eq for the steric factor.

Answer 21

Steric factor p is the fraction of sufficiently energetic collisions which lead to reaction

p = A_experimental/A_{coll theory}

normally p <1, but p >1 when molecules interact over larger distances than predicted by gas kinetic theory