For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Solution and Explanation
For a first order reaction, the time required for 99% completion is
\(t_1 = \frac {2.303}{k} log \ \frac {100}{100-99}\)
\(t_1 = \frac {2.303}{k} log \ 100\)
\(t_1 = 2 \times \frac {2.303}{k}\)
For a first order reaction, the time required for 90% completion is
\(t_2 = \frac {2.303}{k} log \ \frac {100}{100-90}\)
\(t_2 = \frac {2.303}{k} log \ 10\)
\(t_2 = \frac {2.303}{k}\)
\(Therefore,\ t_1= 2t_2\)
Hence, the time required for 99% completion of a first order reaction is twice the time required for the completion of 90% of the reaction.
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Concepts Used:
Order of Reaction
The Order of reaction refers to the relationship between the rate of a chemical reaction and the concentration of the species taking part in it. In order to obtain the reaction order, the rate equation of the reaction will given in the question.
Characteristics of the reaction order
- Reaction order represents the number of species whose concentration directly affects the rate of reaction.
- It can be obtained by adding all the exponents of the concentration terms in the rate expression.
- The order of reaction does not depend on the stoichiometric coefficients corresponding to each species in the balanced reaction.
- The reaction order of a chemical reaction is always defined with the help of reactant concentrations and not with product concentrations.
- Integer or a fraction form the value of the order of reaction will be there and it can be zero.