Question

Consider the free radical polymerization of vinyl acetate with a thermally initiated kinetic mechanism. If the initial concentration of the vinyl acetate is 0.1 mol/L and the polymerization is carried out at 60°C, with a rate constant of 2.5 x 10^-3 L/(mol·s) for the propagation step and an activation energy of 20 kJ/mol, what is the half-life of the reaction?

Answer 1

To determine the half-life of the reaction, we need to consider the rate constant  k  for the decomposition of the initiator, which is the rate-determining step in a free radical polymerization. The rate constant for the propagation step  kp  is given, but we need to find the rate constant for the decomposition of the initiator  k  to calculate the half-life.We can use the Arrhenius equation to find the rate constant for the decomposition of the initiator:k = A * exp -Ea /  R * T  where:k = rate constant for the decomposition of the initiatorA = pre-exponential factor  frequency factor Ea = activation energy  20 kJ/mol = 20,000 J/mol R = gas constant  8.314 J/ molK  T = temperature in Kelvin  60C = 333.15 K Unfortunately, we don't have the value of A. However, we can use the given rate constant for the propagation step  kp  as an approximation for the rate constant for the decomposition of the initiator  k , since they are often of the same order of magnitude in free radical polymerizations.So, we will assume k  kp = 2.5 x 10^-3 L/ mols .Now, we can use the first-order reaction equation to find the half-life:t1/2 = ln 2  / kwhere:t1/2 = half-life of the reactionk = rate constant for the decomposition of the initiator  2.5 x 10^-3 L/ mols  ln 2  = natural logarithm of 2  approximately 0.693 t1/2 = 0.693 /  2.5 x 10^-3 L/ mols  t1/2  277.2 sThe half-life of the reaction is approximately 277.2 seconds.