Chain reaction and rate of formation of KBr

Chain reaction

The chemical reaction In which different intermediate products are formed which successive other products in the series of steps is called chain reaction. Chain reaction are initiated by primary process in reaction is initiated by light, it is called photochemical chain reaction

Let us take an example of photochemical reaction between Hydrogen and Bromide (Br₂)

Chain reaction

H₂and Br₂$$\xrightarrow{\nu}$$ 2HBr

The reaction is assumed to take place in following of steps.

Step 1: Br₂$$\xrightarrow{\nu}$$2 Br^•chain initiation step

Step 2:Br^•+ H₂$$\xrightarrow{k_2}$$2 HBr + H^•

Step 3: H^•+Br₂ $$\xrightarrow{k_3}$$HBr +Br^•

Step 4:H^•+HBr$$\xrightarrow{k_4}$$ H₂+Br^•

Step 5:Br^•+Br^•$$\xrightarrow{k_5}$$ Br₂(chain termination step)

Here, step 2,3 and 4 are chain propagation step.

Rate of formation of HBr

$$\frac{d[HBr]}{dt}$$ = k₂[Br• ][H₂] + k₃[H•][Br₂] - k₄[H^•][HBr]..........(i)

The hydrogen free radical and bromine free radical are intermediate species. These exists for short steady state principle can be applied. According to this principle, for short lived chemical species, the rate of formation is equal to rate of decomposition. This principle can be applied for both hydrogne free radical and bromine free radical.

Rate of formaiton of [H^•], $$\frac{d[H•]}{dt}$$ = k₂[Br•] [H₂] - k₃ [H•] [Br₂] - k₄[H^•][HBr]

0 = k₂[Br•] [H₂] - k₃ [H•] [Br₂] - k₄[H^•][HBr].........(2)

Rate of formation of [Br•], $$\frac{d[Br•]}{dt}$$ = k₁[Br₂] - k₂[Br•][H₂] + k₃[H•][Br₂] + k₄[H^•][HBr] - k₅[Br•]²

0 = k₁[Br₂] -k₂[Br•] [H₂] +k₃ [H•] [Br₂] +k₄[H^•][HBr] - k₅[Br•]²...........(3)

Equation (2) + Equation (3)

k₂[Br•] [H₂] - k₃ [H•] [Br₂] - k₄[H^•][HBr] +k₁[Br₂] -k₂[Br•][H₂] +k₃ [H•] [Br₂] +k₄[H^•][HBr] - k₅[Br•]²= 0

or,k₁[Br₂] -k₅[Br•]²= 0

or,k₁[Br₂] = k₅[Br•]²

or,[Br•]^{2 =} $$\frac{k_1}{k_5}$$[Br₂]

or,$$[Br•] = (\sqrt(\frac{k_1}{k_5} \times [Br_2])...........(4)$$

Putting the value of [Br•] in equation (2)

k₂($$\frac{k_1}{k_5}$$ [Br₂])^{$$\frac{1}{2}$$}[H₂] - k₃ [H•] [Br₂] -k₄[H^•][HBr] = 0

k₃ [H•] [Br₂] + k₄[H^•][HBr]= k₂$$\frac{k_1}{k_5}$$^{$$\frac{1}{2}$$}[Br₂]^{$$\frac{1}{2}$$}[H₂]

[H•]k₃ [Br₂] +k₄[HBr] = k₂$$\frac{k_1}{k_5}$$^{$$\frac{1}{2}$$}[Br₂]^{$$\frac{1}{2}$$}[H₂]

$$[H•] = \frac{k_2 {k_1/k_5}^{1/2} [Br_2]^{1/2} [H_2]}{k_3 [Br_2]+k_4 [HBr]}$$

Putting the value of[Br•] and [H•] in equation (1) then,

$$\frac{d[HBr]}{dt} = k_2 (\frac{k_1}{k_5})^{1/2} [Br_2]^{1/2} [H_2] + k_3 \frac{k_2 {k_1/k_5}^{1/2} [Br_2]^{1/2} [H_2]}{k_3 [Br_2]+k_4 [HBr]} - k_4 (\frac{k_2 (k_1/k_5)^{1/2} [H_2] [Br_2]^{1/2}) [HBr]}{k_3 [Br_2] + k_4 [HBr]}$$

$$= \frac{k_2k_3(k_1/k_5)^{1/2} [Br_2]^{1/2}[Br_2]^{1/2} [H_2] [Br_2]}{k_3 [Br_2] + k_4 [HBr]}$$

$$ = \frac{2k_2 k_3 (k_1/k_5)^{1/2} [H_2] [Br_2]^{1/2}}{\frac{k_3[Br_2]}{k_3[Br_2]} + \frac{k_4[HBr]}{k_3[Br_2]}}$$

$$= \frac{2 k_2 (\sqrt(\frac{k_1}{k_5}) [H_2] [Br_2]^{1/2}}{1 + \frac{k_4}{k_5} \frac{[HBr]}{Br_2}}$$

Kinetics of polymerization

The process of formation of a long chain of high molecular weight by the combination of repeating unit having lower molecular weight with or without loss of small molecules such as H₂O, NH₃and HCl etc is called polymerization and the product formed by this process is called polymer. The repeating basic unit is called monomer. Hence, monomer under repeatation may be addition polymerisation or condensation polymerisation.

Polymerisation

1. Addition polymerisation

• Styrene
• Ethene
• Vinyl chloride

2. Condensation polyrisation

• Polystyrene
• Polyethene
• Polyvinyl chloride
• Teflon

Condensation polymerisation (Nylon 6,6)

• In the condensation polymerisation, there is loss of small molecules such as H₂O, NH₃, HCl etc
• The molecular mass of condensation polymer is not the integral multiple of molecular mass of monomer.
• The polymerisation takes place stepwise. Hence, it is also called step-growth polymerisation.

Kinetics of condensation polymerisation

Let us take an example of condensation polymerisation reaction.

The rate of polymerisation depends upon functional group rather than in the following example.

The rate of chemical reaction depends upon the active functional group rather than the length of the chain. In group are -COOH and -OH, Hence, rate of reaction depends upo two functional group. Hence, above reaction can be considered as second order reaction int he concentration of -COOH and -OH.

- $$\frac{d}{dt}$$ [-COOH] = k [-COOH]{-OH]............(1)

SInce, concentration of [-COOH] = [-OH] let it be c

Now the rate law equation becones

- $$\frac{dc}{dt}$$ = kc²...........(2)

- $$\frac{dc}{dc^2}$$ = kdt

Now integrating,

- $$\int$$ $$\frac{dc}{c^2}$$ = k$$\int$$ dt

-(-$$\frac{1}{c}$$ = kt + constant.........(3)

when t = 0, c = c_owhere c_ois the initial concentration of -COOH or -OH

∴ constant = $$\frac{1}{c_o}$$.........(4)

y = mx + c

If a graph of $$\frac{1}{c}$$ vs timw ia plotted then a straighr line with a slope of rate constnat (k) is obtained.

References