the half life period of first order reaction is 600s. what% of a remains after 30 minutes
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The half life of a first order reaction is given by the equation,
t(1/2) = 0.693/k
Where “k” is the rate constant.
From the equation, we can calculate the rate constant.
60 = 0.693/k
k = 0.693/60 = 0.01155 min^(-1)
Now, use the equation
k= (2.303/t)log(initial conc./final conc.)
After consuming 90% of the reactant, the final concentration will be 10% of the initial concentration.
Assume, initial concentration is x
Final concentration is 10% of x = (10/100)x = 0.1x
Therefore
Initial conc./final conc. = x/0.1x = 10
Substitute this in the above formula
0.01155 = (2.303/t)log(10)
0.01155 = 2.303/t
t = 199.4 minutes.
Therefore it takes 199.4 minutes to consume 90% of the reactant.
t(1/2) = 0.693/k
Where “k” is the rate constant.
From the equation, we can calculate the rate constant.
60 = 0.693/k
k = 0.693/60 = 0.01155 min^(-1)
Now, use the equation
k= (2.303/t)log(initial conc./final conc.)
After consuming 90% of the reactant, the final concentration will be 10% of the initial concentration.
Assume, initial concentration is x
Final concentration is 10% of x = (10/100)x = 0.1x
Therefore
Initial conc./final conc. = x/0.1x = 10
Substitute this in the above formula
0.01155 = (2.303/t)log(10)
0.01155 = 2.303/t
t = 199.4 minutes.
Therefore it takes 199.4 minutes to consume 90% of the reactant.
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