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What is the concentration of $A$ at time $45\;s$ if $[A]_o = 1\;M$, $[B]_o = 45\;M$, and 2nd order rate constant is $0.6\;M^{-1}s^{-1}$?

$\begin{array}{1 1} 1.88 \times 10^{-12}\;M \\ 3.76 \times 10^{-12}\;M \\ 0.94 \times 10^{-12}\;M \\ 4.7 \times 10^{-12}\;M \end{array}$

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Answer: $1.88 \times 10^{-12}\;M$
Given $[A]_o = 1\;M$, $[B]_o = 45\;M$, and 2nd order rate constant is $0.6\;M^{-1}s^{-1}$?
Since $[B]_0 \gt \gt [A]_0$ we can treat this as a pseudo-first order reaction.
We can use the rate equation $[A] = [A]_0 e^{-k't[B]_0}$
$\Rightarrow [A] = 1\;M\;e^{-0.6M^{-1}s^{-1} \times 45\;s \times 45\;M} =1.88 \times 10^{-12}\;M$
answered Jul 25, 2014 by balaji.thirumalai
 

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