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What is the half-life of a reaction with $[A]_o = 109\;M$, $[B]_o = 1\;M$, $k' = 45M^{-1}s^{-1}$ ?

$\begin{array}{1 1} 1.41 \times 10^{-4}\;s \\ 4905\;s \\ 0.141 \times 10^{-4}\;s \\ 490.5\;s \end{array}$
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Answer: $1.41 \times 10^{-4}\;s$
Given: $[A]_o = 109\;M$, $[B]_o = 1\;M$, $k' = 45M^{-1}s^{-1}$
Because $[A]_0 \gt \gt [B]_0$, this is a pseudo first order reaction and we can multiply $k'$ with $[A]_0$ to find $k$.
$\Rightarrow 109\;M \times 45M^{-1}s^{-1} = 4905s^{-1}$
Half life of a pseudo first order reaction is $ t_{1/2} = \large\frac{ln \;0.5}{-k}$
$\Rightarrow t_{1/2} = \large\frac{ln\;0.5}{4905s^{-1}}$$ = 1.41 \times 10^{-4}\;s$
answered Dec 13, 2013 by sreemathi.v
edited Jul 25, 2014 by balaji.thirumalai

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