The gas phase reaction A2(g) + 2 B(g) → 2 AB(g) is accelerated by catalyst C. The overall rate const — Physical Chemistry — Kinetics Chemistry Question
Kinetics in gas phase
The gas phase reaction
A2(g) + 2 B(g) → 2 AB(g)
is accelerated by catalyst C. The overall rate constant is found to increase linearly with the catalyst concentration. Following measurements are done at 400 K with [C] = 0.050 mol·dm–3:
[VISUAL]
| Experiment No | [A2] (mol·dm–3) | [B] (mol·dm–3) | Initial rate (mol·dm–3·s-1) |
|---|---|---|---|
| 1 | 0.010 | 0.10 | 1.600 · 10^-10 |
| 2 | 0.010 | 0.20 | 3.200 · 10^-10 |
| 3 | 0.100 | 0.20 | 1.012 · 10^-9 |
What is the rate law of this reaction?
Model Answer
From exp. 1 and 2:
(3.200 · 10^-10) / (1.600 · 10^-10) = (k [0.01]^x [0.20]^y) / (k [0.01]^x [0.10]^y)
Thus: y = 1
From exp 3 and 2:
(1.012 · 10^-9) / (3.200 · 10^-10) = (k [0.10]^x [0.20]^y) / (k [0.010]^x [0.20]^y)
Thus: x = 0.5
Therefore, the rate of the reaction = k [A2]^(1/2) [B] [C] = koverall [A2]^(1/2) [B]
(Note: The solution text in the source contains a typo writing [B2] instead of [B]: "Therefore, the rate of the reaction = k [A2]^1/2 [B2] [C] = koverall [A2]^1/2 [B2]")
Calculate the numerical value of koverall at 400 K.
Model Answer
koverall = k[C]
Rate = 1.600 · 10^-10 = koverall × 0.01^(1/2) × 0.1
koverall = 1.6 · 10^-8 dm^(3/2) mol^(-1/2) s^-1
For this hypothetical reaction the following mechanism was proposed:
A2(g) ⇌ 2 A(g) (fast equilibrium)
A(g) + B(g) + C(g) → ABC(g) (slow step)
ABC(g) → AB(g) + C(g)
Check that the suggested mechanism gives the equation for the overall reaction.
Model Answer
After multiplying second and third reaction equations by two and adding the three steps, the overall reaction equation is found:
A2(g) + 2 B(g) → 2 AB(g)
Show that the suggested mechanism is consistent with the rate law determined experimentally.
Model Answer
Rate of disappearance of B = 2 * d[B]/dt = k2 [A][B][C]
From the fast equilibrium step: [A]^2 / [A2] = k1 / k-1, which gives [A] = [VISUAL] * [A2]^(1/2)
Rate of disappearance of B = 2 * k2 * [VISUAL] * [A2]^(1/2) * [B][C]
Since the overall rate of the reaction is defined as:
Rate of the reaction = -d[A2]/dt = -1/2 * d[B]/dt = 1/2 * d[AB]/dt
Rate of the reaction = 1/2 * 2 * k2 * [VISUAL] * [A2]^(1/2) * [B][C]
Rate of the reaction = k2 * [VISUAL] * [A2]^(1/2) * [B][C] = koverall * [A2]^(1/2) * [B]
where koverall = k2 * [VISUAL] * [C]
This is consistent with the experimentally determined rate law.
Calculate the dissociation enthalpy of A2 bond using the following information:
• At 400 K, when [A2] is 1.0 · 10–1 mol dm–3, [A] is 4.0 · 10–3 mol dm–3.
• When the first experiment is repeated at 425 K, the initial reaction rate increases to a three-fold value.
• Activation energy of the slowest step is 45.0 kJ.
Model Answer
Since koverall = k2 * [VISUAL] * [C]:
In order to find the dissociation enthalpy of A2 we need to find the equilibrium constant K_eq = k1/k-1 at both 400 K and 425 K.
At 400 K:
K_eq(400) = [A]^2 / [A2] = (4.0 · 10^-3)^2 / 0.10 = 1.6 · 10^-4
From the relationship koverall = 1/2 * k2 * [VISUAL] * [C]:
1.6 · 10^-8 = 1/2 * k2(400) * (1.6 · 10^-4)^(1/2) * 0.050
k2(400) = 5.06 · 10^-5
Using the Arrhenius equation for the slow step at 425 K:
[VISUAL] = Ea / R * (1/T1 - 1/T2)
ln(k2(425) / 5.06 · 10^-5) = (45000 / 8.314) * (1/400 - 1/425)
k2(425) = 1.12 · 10^-4
At 425 K, the rate of the first experiment increases 3-fold, meaning koverall(425) = 3 * koverall(400) = 4.8 · 10^-8:
4.8 · 10^-8 = 1/2 * (1.12 · 10^-4) * (K_eq(425))^(1/2) * 0.050
(K_eq(425))^(1/2) = 1.71 · 10^-2
K_eq(425) = 2.93 · 10^-4
Using the van 't Hoff equation to find the dissociation enthalpy (ΔH_diss) of the A2 bond:
ln(K_eq(425) / K_eq(400)) = (ΔH_diss / R) * (1/T1 - 1/T2)
ln(2.93 · 10^-4 / 1.6 · 10^-4) = (ΔH_diss / 8.314) * (1/400 - 1/425)
0.605 = (ΔH_diss / 8.314) * (1.47 · 10^-4)
ΔH_diss = 34.2 kJ/mol