The observed rate law for a chemical reaction can arise from several different mechanisms. For the r — Physical Chemistry — Kinetics Chemistry Question
Rates and reaction mechanisms
The observed rate law for a chemical reaction can arise from several different mechanisms. For the reaction
H2 + I2 → 2 HI
the observed rate law is
d[H2]/dt = -k[H2][I2]
For a long time it was believed that the above reaction took place as it was written down; that is, it was a bimolecular elementary reaction. It is now considered that several mechanisms compete. Below a certain temperature, two alternative mechanisms have been proposed :
(1) I2 ⇌ 2 I (K : equilibrium constant)
I + I + H2 → 2 HI (rate constant k1)
(2) I2 ⇌ (I2)d (K' : equilibrium constant)
(I2)d + H2 → 2 HI (rate constant k1')
where (I2)d represents a dissociative state of I2. The first step in each mechanism is fast and the second slow.
Show that both mechanisms are consistent with the observed rate law.
Model Answer
Mechanism 1:
d[HI]/dt = k1 [I]^2 [H2]
Since the first step is fast, there is a pre-equilibrium:
K = [I]^2 / [I2]
∴ d[HI]/dt = k1 K [I2][H2] = k [I2][H2]
Mechanism 2 :
d[HI]/dt = k1' [(I2)d] [H2]
K' = [(I2)d] / [I2]
d[HI]/dt = k1' K' [I2][H2] = k [I2][H2]
Both mechanisms are consistent with the observed rate law.
The values of the rate constant k for the reaction at two different temperatures are given in the table :
Determine the activation energy Ea.
Model Answer
i. k = A e^(-Ea/RT)
ln(k2/k1) = Ea/R (1/T1 - 1/T2)
With the given numerical values,
Ea = 170 kJ mol^-1
The bond dissociation energy of I2 is 151 kJ mol–1. Justify why the second step in each mechanism is rate determining.
Model Answer
The activation energy is greater than the bond dissociation energy of I2. Hence the second step is rate determining in both the mechanisms.
The change in internal energy (ΔU) for the reaction is –8.2 kJ mol–1. Determine the activation energy for the reverse reaction.
Model Answer
The activation energy Ea' for the reverse reaction is
Ea' = Ea - ΔU = 170 + 8.2 = 178.2 kJ mol^-1
The activation energy for a reaction can even be negative. An example is the gas phase recombination of iodine atoms in the presence of argon
I + I + Ar → I2 + Ar ,
whose activation energy is about –6 kJ mol–1.
One of the proposed mechanisms of this reaction is :
I + Ar + Ar ⇌ IAr + Ar (K'' : equilibrium constant)
Ar + I → I2 + Ar (rate constant k3)
where IAr is a very loosely bound species.
Assume that the second step is the rate determining and obtain the rate law for the reaction.
Model Answer
d[I2]/dt = k3 [IAr] [I]
K'' = [IAr] / ([I][Ar])
d[I2]/dt = k3 K'' [I]^2 [Ar]^2 = k [I]^2 [Ar]^2
Give a possible explanation of why the activation energy for the iodine recombination is negative.
Model Answer
A possible reason why this is negative is that Ea3 is positive and less in magnitude than |ΔH°|, while ΔH° is negative.
k = k3 K''
k = A e^(-Ea3/RT) e^(-ΔG°/RT)
Since ΔG° = ΔH° - TΔS°
k = (A e^(ΔS°/R)) e^(-(Ea3 + ΔH°)/RT)
The activation energy for the overall reaction is Ea3 + ΔH°.