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VITEEE 2014 Paper

Section: Chemistry

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Question : 53 of 120

Marks: +1, -0

The half-life of two samples are 0.1 and 0.8 s. Their respective concentration are 400 and 50 respectively. The order of the reaction is

0
2
1
4

Solution:

It is known,

\frac{\left(t_{\frac{1}{2}}\right)_1}{\left(t_{\frac{1}{2}}\right)_2}

=

\left[\frac{a_2}{a_1}\right]^{(n-1)}

Here, n = order of the reaction

Given ,

\left(t_{\frac{1}{2}}\right)_1

= 0.1 s ,

a_1

400

\left(t_{\frac{1}{2}}\right)_2

= 0.8 s ,

a_2

= 50

On putting the values,

\frac{0.1}{0.8}

=

\left[\frac{50}{400}\right]^{(n-1)}

Taking log on both sides

log

\frac{0.1}{.8}

= (n - 1) log

\frac{50}{400}

log

\frac{1}{8}

= (n - 1) log

\frac{1}{8}

n - 1 = 1 ⇒ n = 2

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