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At $150^oC$ the decomposition of acetaldehyde to methane is first order reaction. If rate constant for reaction at $750^oC$ is $0.05m^{-1}$, how long does it take a concentration of $0.04\;mol\;L^{-1}$ of acetaldehyde to reduce concentration of $0.03\;mol\;L^{-1}$?

$\begin{array}{1 1}(a)\;2.75min&(b)\;3.75min\\(c)\;4.75min&(d)\;5.75min\end{array}$

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For first order kinetics
$K\times t=ln\large\frac{a}{a-x}$
$a$=initial concentration=0.04
$a-x$=reactant concentration at t time=0.03
$0.05\times t=ln\big(\large\frac{0.04}{0.03}\big)$
$t=5.75min$
Hence (d) is the correct answer.
answered Dec 18, 2013 by sreemathi.v
 

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