Test Index

Chemical Kinetics Part 3

Section: Chemistry

Question : 4 of 24

Marks: +1, -0

(t_{1\/2})

[A]_0

Solution:

For zero order

t_{1 \/ 2}=\;{A_0}/{2 k}

y=m x

Straight line starting from origin
The given graph follow zero order kinetic

t_{1 \/ 2} ∝ A_0

\;{ (t_{1 \/ 2}) _1}/{ (t_{1 \/ 2}) _2}=\;{ (A_0) _1}/{ (A_0) _2}

\;{240}/{x}=\;{4 × 10^{-3}}/{1.5 × 10^{-3}}

x=\;{240 × 1.5}/{4}

x=90 {\ s}

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