For a first order reaction A-_→ 2B + C. It was found that at the end

1.	For a first order reaction A-_→ 2B + C. It was found that at the end of 10 minutes from the start, thetotal optical rotation of the system was 60° and when the reaction is complete, it was 180°. The B andC are only optically active and intially only A was taken(i) What is the rate constant of the above reaction (in hour)?(ii) At what time (in minute) from the start, total optical rotation will be 90°(Take log2 0.3, log3 0.48, log7 0.85, In10 2.3)
Answer» A => 2B + C p 0 0 (initial) (at t=0) P-Po 2Po Po (final) (at t =10) initial total pressure = P = 90mm final total pressure = 180mm from eq adding all presures we get total pressure = P + 2Po P + 2Po= 180 90 + 2Po= 180 Po= 45 for A , initially pressure = P= 90 final pressure = P-Po= 90-45 = 45 rate constant (k) = 2.303 log(Pi/Pf) / t = 2.303 log90/45 / t =(2.303)log2/10 k = 0.069 min-1 this value of rate constant is in min-1, 1min = 60sec so K = 0.069/60 sec-1

For a first order reaction A-_→ 2B + C. It was found that at the end of 10 minutes from the start, thetotal optical rotation of the system was 60° and when the reaction is complete, it was 180°. The B andC are only optically active and intially only A was taken(i) What is the rate constant of the above reaction (in hour)?(ii) At what time (in minute) from the start, total optical rotation will be 90°(Take log2 0.3, log3 0.48, log7 0.85, In10 2.3)

Answer»

A => 2B + C

p 0 0 (initial) (at t=0)

P-Po 2Po Po (final) (at t =10)

initial total pressure = P = 90mm

final total pressure = 180mm

from eq adding all presures we get total pressure = P + 2Po

P + 2Po= 180

90 + 2Po= 180

Po= 45

for A , initially pressure = P= 90

final pressure = P-Po= 90-45 = 45

rate constant (k) = 2.303 log(Pi/Pf) / t

= 2.303 log90/45 / t

=(2.303)log2/10

k = 0.069 min-1

this value of rate constant is in min-1, 1min = 60sec so

K = 0.069/60 sec-1