for a first order polymerisation rxn: nA(g)−>An(g), occuring at const volume n temp.

the half life of polymerisation of ‘A’ is 20min. If the total press. of system is 2atm at t=0 and 1.2atm at t=20min,then the value of ‘n’ is:

To determine the value of 'n' in the first-order polymerization reaction \( nA(g) \rightarrow A_n(g) \), we can use the information provided about the half-life and the change in pressure over time. Let's break this down step by step.

Understanding the Reaction Dynamics

In a first-order polymerization reaction, the rate of reaction is proportional to the concentration of the reactant. For our reaction, the initial concentration of the monomer 'A' can be related to the total pressure of the system, since pressure is directly proportional to the number of moles of gas present.

Initial Conditions

At time \( t = 0 \), the total pressure \( P_0 \) is 2 atm. This pressure corresponds to the concentration of the monomer 'A' before any polymerization occurs. As the reaction proceeds, some of the monomer 'A' is converted into the polymer \( A_n \), which does not contribute to the pressure in the same way because it is a larger molecule.

Pressure Change Over Time

After 20 minutes, the pressure drops to 1.2 atm. This decrease in pressure indicates that some of the monomer has reacted to form the polymer. The change in pressure can be calculated as follows:

• Initial pressure \( P_0 = 2 \, \text{atm} \)
• Pressure after 20 minutes \( P_{20} = 1.2 \, \text{atm} \)
• Change in pressure \( \Delta P = P_0 - P_{20} = 2 - 1.2 = 0.8 \, \text{atm} \)

Relating Pressure Change to Moles of A

In a first-order reaction, the half-life \( t_{1/2} \) is given by the formula:

t_1/2 = 0.693 / k

Where \( k \) is the rate constant. Given that the half-life is 20 minutes, we can find \( k \) as follows:

k = 0.693 / 20 \, \text{min} = 0.03465 \, \text{min}^{-1}

Calculating the Extent of Reaction

For a first-order reaction, the relationship between the concentration of reactants and time can be expressed as:

ln(P_0 / P_t) = kt

Substituting the known values:

ln(2 / 1.2) = 0.03465 \times 20

Calculating the left side:

ln(1.6667) \approx 0.511

Calculating the right side:

0.03465 \times 20 \approx 0.693

Since the left side is approximately equal to the right side, this confirms our calculations are consistent.

Finding the Value of n

The decrease in pressure corresponds to the amount of monomer that has reacted. Each mole of \( A \) that reacts produces \( n \) moles of \( A_n \). Therefore, if \( x \) moles of \( A \) have reacted, the change in pressure can be expressed as:

ΔP = x(1 - 1/n)

From our previous calculations, we know that \( ΔP = 0.8 \, \text{atm} \). Thus, we can set up the equation:

0.8 = x(1 - 1/n)

Relating x to Initial Pressure

Initially, we had 2 atm, and after 20 minutes, we have 1.2 atm, which means:

x = (2 - 1.2) = 0.8 \, \text{atm}

Substituting this value into our earlier equation:

0.8 = 0.8(1 - 1/n)

Dividing both sides by 0.8 gives:

1 = 1 - 1/n

Rearranging this gives:

1/n = 0

Thus, we find:

n = 1

Final Thoughts

In conclusion, the value of 'n' for the polymerization reaction is 1, indicating that the reaction does not lead to the formation of a polymer but rather that the monomer remains unchanged. This is a unique case in polymer chemistry, highlighting the importance of understanding reaction kinetics and the relationship between pressure and concentration in gaseous systems.