The catalytic decomposition of isopropanol on the surface of a V2O5 catalyst, leading to the product — Physical Chemistry — Kinetics Chemistry Question
Catalytic Decomposition of Isopropanol
The catalytic decomposition of isopropanol on the surface of a V2O5 catalyst, leading to the products in the scheme, satisfies a first order kinetic equation.
Five seconds after initiation of the reaction at 590 K, the concentrations of the products in the reaction mixture are:
cC = 28.2 mmol dm-3
cB = 7.8 mmol dm-3
cD = 1.8 mmol dm-3
After the reaction has been completed, the concentration of the product C is 94 mmol dm-3.
Calculate the initial concentration of isopropanol, c0A.
Model Answer
cA0 = cC∞ = 94.0 mmol dm-3
Calculate the velocity constant, k, for the reaction.
Model Answer
k = 1/t × 2.303 log ( cA0 / cA ) = 1/5 × 2.303 log (94.0 / (94.0 - 28.2 - 7.8 - 1.8))
k = 1/5 × 2.303 log (94.0 / 56.2) = 1/5 × 2.303 × 0.223 = 0.0983 s-1
Calculate the half-life of the reaction, t1/2.
Model Answer
t1/2 = 2.303 log 2 / k = (2.303 × 0.301) / 0.0983 = 7.05 s
Calculate the velocity constants for the partial reactions, k1, k2 and k3.
Model Answer
v1 = ∆cB / ∆t = k1 cA
v2 = ∆cC / ∆t = k2 cA
v3 = ∆cD / ∆t = k3 cA
v = v1 + v2 + v3 = k cA
(1) k1 + k2 + k3 = k = 0.0983 s-1
∆cB / ∆cC = cB / cC = k1 / k2 = 7.8 / 8.3 = 0.940
∆cB / ∆cD = cB / cD = k1 / k3 = 7.8 / 1.8 = 4.33
From equations (1) – (3):
k1 = 0.0428 s-1
k2 = 0.0455 s-1
k3 = 0.00988 s-1
Calculate the concentrations of the compounds A, B, C and D at time t = t1/2.
Model Answer
At t = τ 1/2 = 7.05 s
(4) cA = cA0 / 2 = 47.0 mmol dm-3
cA0 = cA + cB + cC + cD = 94.0 mmol dm-3
From equations (2) – (4):
cB = 10.0 mmol dm-3
cC = 10.7 mmol dm-3
cD = 2.32 mmol dm-3