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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

```
9 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

```
9 years ago
```							Dear student.,
Write the rate of reation in terms of pressure
r=(2.303log P/Po)/time

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```
9 years ago
```							Dear Student,Please find below the solution to your problem.A =>      2B + Cp 0 0 (initial) (at t=0)P-Po 2Po Po (final) (at t =10)initial total pressure = P = 90mmfinal total pressure = 180mmfrom eq adding all presures we get total pressure = P + 2PoP + 2Po = 18090 + 2Po = 180 Po = 45for A , initially pressure = P= 90 final pressure = P-Po = 90-45 = 45rate constant (k) = 2.303 log(Pi/Pf) / t            = 2.303 log90/45 / t =(2.303)log2/10 k = 0.069 min-1this value of rate constant is in min-1 , 1min = 60sec so K = 0.069/60 sec-1 = 1.15*10-3 sec-1Thanks and Regards
```
2 months ago
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