Bissoy.com
Doctors Medicines MCQs Q&A

A first order gaseous reactions has K = 1.5 x 10-6 sec-1 at 200°C. If the reaction is allowed to run for 10 hour, what percentage of initial concentration would have changed into products. What is the half life period of reaction?

6 views

2 Answers

Given

k = 1.5 x 10-6 s-1

t = \(\frac{2.303}k\) log\(\frac{No}{No-N}\) 

Where No = initial concentration

No - N = concentration After 10 hour.

N = concentration which reacted.

\(\therefore\) 10 x 60 x 60  = \(\frac{2.303}{1.5\times10^{-6}}log\frac{No}{No-N}\) 

log\(\frac{No}{No-N}\) = \(\frac{10\times60\times60\times60\times1.5\times10^{-6}}{2.303}\) 

log\(\frac{No}{No-N}\) = 0.02345

or log \(\frac{N_o-N}{N_o}=-0.2345\)

taking antilog on both side---

\(\frac{N_o-N}{N_o}=0.9474\)

1 - \(\frac{N}{N_o}=0.9474\)

\(\frac{N}{N_o}=1-0.9474\)

\(\frac{N}{N_o}=0.0527\)

\(\therefore\) percentage of initial concentration change into product = \(\frac{N}{N_o}\times100\)

 = 0.0527 x 100

 = 5.27 %

t1/2 = \(\frac{0.693}k\) 

 = \(\frac{0.693}{1.5\times10^{-6}}\) 

 = 0.462 x 106

 t1/2 = 462000 sec

6 views

Given

k = 1.5 x 10-6 s-1

t = \(\frac{2.303}k\) log\(\frac{No}{No-N}\) 

Where No = initial concentration

No - N = concentration After 10 hour.

N = concentration which reacted.

\(\therefore\) 10 x 60 x 60  = \(\frac{2.303}{1.5\times10^{-6}}log\frac{No}{No-N}\) 

log\(\frac{No}{No-N}\) = \(\frac{10\times60\times60\times60\times1.5\times10^{-6}}{2.303}\) 

log\(\frac{No}{No-N}\) = 0.02345

or log \(\frac{N_o-N}{N_o}=-0.2345\)

taking antilog on both side---

\(\frac{N_o-N}{N_o}=0.9474\)

1 - \(\frac{N}{N_o}=0.9474\)

\(\frac{N}{N_o}=1-0.9474\)

\(\frac{N}{N_o}=0.0527\)

\(\therefore\) percentage of initial concentration change into product = \(\frac{N}{N_o}\times100\)

 = 0.0527 x 100

 = 5.27 %

t1/2 = \(\frac{0.693}k\) 

 = \(\frac{0.693}{1.5\times10^{-6}}\) 

 = 0.462 x 106

 t1/2 = 462000 sec

6 views

Related Questions