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Chemical Kinetics

Question

Show that the time required for the completion of 3/4th reaction of first order is twice the time required for completion of 1/2 of the reaction.

Solution

Since it is first order, thus

$t = \frac{2.303}{k} \log \frac{a}{a - x} also x = 3 a / 4 so the equation is t_{3 / 4} = \frac{2.303}{k} \log \frac{a}{a - 3 a / 4} . . . . . 1 for x = 1 a / 2 t_{1 / 2} = \frac{2.303}{k} \log \frac{a}{a - a / 2} . . . . . . 2 divide the 1 by 2, we get \frac{t_{3 / 4}}{t_{1 / 2}} = \frac{\frac{2.303}{k} \log \frac{a}{a - 3 a / 4}}{\frac{2.303}{k} \log \frac{a}{a - a / 2}} thus we get \frac{t_{3 / 4}}{t_{1 / 2}} = \frac{\log 4}{\log 2} \frac{t_{3 / 4}}{t_{1 / 2}} = \frac{2 \log 2}{\log 2} \frac{t_{3 / 4}}{t_{1 / 2}} = 2 t_{3 / 4} = 2 t_{1 / 2}$

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Chemical Kinetics

Show that the time required for the completion of 3/4th reaction of first order is twice the time required for completion of 1/2 of the reaction.

Some More Questions From Chemical Kinetics Chapter

A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of the reaction.

The rate law for the decomposition of N₂O₅ is rate = k[N₂O₅]. What is the significance of k in this equation?

Define activation energy of a reaction.

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For Daily Free Study Material Join wiredfaculty

Chemical Kinetics

Show that the time required for the completion of 3/4th reaction of first order is twice the time required for completion of 1/2 of the reaction.

Some More Questions From Chemical Kinetics Chapter

A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of the reaction.

The rate law for the decomposition of N2O5 is rate = k[N2O5]. What is the significance of k in this equation?

Define activation energy of a reaction.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

The rate law for the decomposition of N₂O₅ is rate = k[N₂O₅]. What is the significance of k in this equation?