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for the first order reaction A(g) ------: 2B(g) +C(g) the initial pressure is Pa= 90 mm hg and pressure after 10 min. is 180 mm of hg. find the rate constant.ans is 1.15*10^-3


for the first order reaction


A(g) ------: 2B(g) +C(g)


the initial pressure is Pa= 90 mm hg and pressure after 10 min. is 180 mm of hg. find the rate constant.ans is 1.15*10^-3


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3 Answers

vikas askiitian expert
509 Points
13 years ago

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  = 1.15*10-3 sec-1

approve if u like my ans

SAGAR SINGH - IIT DELHI
878 Points
13 years ago

Dear student.,

Write the rate of reation in terms of pressure

r=(2.303log P/Po)/time

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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

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 = 1.15*10-3 sec-1

Thanks and Regards

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