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The rate of a first-order reaction is 0.04 mol l^-1 s^-1 at 10 seconds and 0.03 mol l^-1 s^-1 at 20 seconds after initiation of the reaction The half life period of the reaction is

amit singh , 8 Years ago
Grade 11
anser 3 Answers
Vikas TU

Last Activity: 8 Years ago

 U have to just put the values into the formulae for the first order eqn. that is:
K = (2.303/t2 –  t1)log(Q1/Q2)
   =(2.303/(10)llog(0.04/0.03)
After solving,
K = 0.02876 /min
thalf = 0.693/k = > 24 min approx.

Garima Shrivastava

Last Activity: 6 Years ago

Solution is all correct above just 
 
K = (2.303/t2 –  t1)log(Q1/Q2)
   =(2.303/(10)llog(0.04/0.03)
After solving,
 
K= 2.6 / second*
And time = 24 seconds* approximately

Garima Shrivastava

Last Activity: 6 Years ago

K value went wrong. Uploading again. Sorry
Solution is all correct above just 

 

 

K = (2.303/t2 –  t1)log(Q1/Q2)


   =(2.303/(10)llog(0.04/0.03)

After solving,

 

K=  0.02 second*

 

And time = 24 seconds* approximately

 

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