[13%] Ethene, C2H4, can react in the gas phase in the presence of radicals R to form polyethylene as — Kinetics and Thermodynamics Chemistry Question
Problem Context
[13%] Ethene, C2H4, can react in the gas phase in the presence of radicals R to form polyethylene as shown in the equation below. Here n is the degree of polymerization. The forward reaction is second-order while the reverse reaction is first-order. The values of these rate constants are independent of the degree of polymerization n and the identity of R.
A sample of polyethylene has an average degree of polymerization n = 1200. How many polymer chains are present in 1.0 g of this material?
Model Answer
M = 1200(28.05 g mol-1) = 3370 g mol-1
Thus 1.0 g polymer = 3.0 × 10-5 mol = 1.8 × 1019 molecules
Calculate ∆H° and ∆S° for the polymerization reaction.
Model Answer
Keq = kf/kr, so ln(Keq) = ln(kf) – ln(kr) = 10500(1/T) – 16.5
Since ∆G° = –RTln(Keq) = ∆H° – T∆S°, ln(Keq) = –(∆H°/R)(1/T) + (∆S°/R)
∆H° = –10500R = –87.3 kJ mol-1
∆S° = –16.5R = –137 J mol-1 K-1
The bond dissociation enthalpy (BDE) for a typical carbon-carbon single bond is 345 kJ mol-1. From the data given, what is the BDE of the carbon-carbon double bond in ethene?
Model Answer
∆H°rxn = {BDE of C=C} – 2×{BDE of C–C} = –87.3 kJ mol-1
{BDE of C=C} – 2×(345 kJ mol-1) = –87.3 kJ mol-1
{BDE of C=C} = 603 kJ mol-1
Ethene is charged to a fixed vessel at 25 bar and 720 K. Traces of radical are then added to initiate polymerization. What is the percent conversion of ethene into polymer at equilibrium under these conditions?
Model Answer
Since the properties of the shorter and longer polymer radicals are the same, their concentrations are effectively equal, so Kp = 1/PC2H4.
Kp = kf/kr = 0.147 at 720 K
So at equilibrium PC2H4 = 6.8 bar. Thus 6.8/25 × 100% = 27% unreacted ethene, 73% incorporation into polymer.
In the presence of a catalyst for the polymerization reaction, the forward rate constant as a function of temperature is ln(kf) = –3050(1/T) + 21.0. By what factor does the catalyst accelerate the rate of the forward reaction at 500 K?
Model Answer
From the equations:
kf (uncatalyzed, 500 K) = 1310 bar-1 s-1
kf (catalyzed, 500 K) = 2.96 × 106 bar-1 s-1
kcat/kuncat = 2300
The catalyzed reaction is 2300 times faster than the uncatalyzed one.
By what factor does the catalyst change the rate of the reverse reaction at 500 K?
Model Answer
Since addition of a catalyst cannot change Keq, the reverse reaction must also be accelerated by a factor of 2300 by the catalyst.