The following initiators are available for batch polymerization of methyl methacrylate:

Initiator t$1/2,70\backslash$deg C $($h$)$

Azobisisobutyronitrile$=4\mathrm{.}8$

Benzoyl peroxide$=7\mathrm{.}3$

Acetyl peroxide$=8\mathrm{.}1$

Lauryl peroxide $=3\mathrm{.}5$

The heat transfer capacity of the cooling system is $250$ kW$/$m$3.$ A $70\%$ conversion must be reached within $1,0$ h$.$ The rate constant for propagation and termination is $1000$ dm$3/($mol$*$s$)$ and $10^7$ dm$3/($mol$*$s$),$ respectively. Initiator efficiency $($f$)$ is $0\mathrm{.}65$

Choose the most suitable initiator and initiator concentration so that the following requirements are met:

Initial monomer concentration $([$M$]0)$ is $5$ mol$/$dm$3.$ Reaction heat of methyl methacrylate polymerization is $100$ kJ$/$mol$.$ You can assume that the initiators have same price.

Draw plots showing monomer and initiator concentration versus time.

HINTS are below

All the equations you need can be found in the lecture on Polymer Reaction Engineering, chapter $4$ 'Radical polymerization'.

First, you need to calculate the initial concentration of each initiator and see which one is the best. As all the initiators have the same prices, the best one is the one with the less initial concentration. Then plot initiator $[1{]}_0$ and monomer $M$ concentration versus time, and at

the end, calculate the heat of the reaction because it needs to be lower than the heat transfer capacity of the cooling system

Part A$.$ Finding an equation for monomer concentration $[$M$].$

Using the rate of polymerization equations, you need to extract the rate of polymerization

equation:

$R_p=k_p[M](\frac{R_i}{2k}{)}^{\frac{1}{2}}$

$k_d$ is the rate constant of initiation $(s^{-1}).$

Substitute $R_i$ with an appropriate equation to get:

$R_p=k_p[M](\frac{f{k}_d[I]}{k_t}{)}^{\frac{1}{2}}$

Substitute $[$I$]$ with an appropriate equation; then use the rate of polymerization equation

$(-\frac{d[M]}{dt}=R_P)$ to get

$($d$[$M$])/($dt$)=$ Kp$\{($f$.$kd$.[$I$]$o$.$e$^(-$kd$.$t$))/$kt$\}^(1/2)$ dt

Do Integration and plot $M$ versus time.

Part B$.$ finding the initial concentration, $[$I${]}_0$

After integrating from the above equation, substitute $M$ with the following relation and

act $[I{]}_0$ for all initiators to find the best one

$[M]=[M{]}_0(1-p)or\frac{M}{[M{]}_0}=(1-p)$

ere $p$ is the extent of the reaction and $[M{]}_0$ is the initial concentration of the monomer

$\{$:$\frac{1}{d}{m}^3).$

Finally, plot $[$I$]$ versus time.

$[I]=[I{]}_0{e}^{-kdt}$

Part C$.$ The released heat $($heat of reaction$)$

The heat produced by the reaction to reach the conversion of monomer at $10$ minutes is

calculated by this equation :

$\frac{Q}{t}=\frac{H[M{]}_0{x}_t}{t}$

Where $\frac{Q}{t}$ is the heat formed $(\frac{kJ}{m^3.s}),H$ is the reaction heat of methyl methacrylate polymerization $(k\frac{J}{m}ol),x_t$ is the conversion of the monomer at time $t$ and $t$ is the time $($s$).$

The conversion of the monomer at time $t$ is calculated by:

$x_t=\frac{[M{]}_0-[M]}{[M{]}_0}$