Q. The slope of $lnK$ vs $\frac{1}{ T}$ was found to be $-40$ if' $'T'$ is in kelvin. The value of $lnμ_{160 K}$ will be : $\mu _{T}$ represents temperature coefficient at $T$ Kelvin $\left.$

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A

$\frac{1}{17}$

B

$\frac{1}{68}$

C

$68$

D

$17$

Solution:

$\ell nK=\ell nA-\frac{Ea}{RT}\&\mu =\frac{K_{170}}{ K_{160}}$
slope $=\frac{Ea}{R}$
$\frac{- Ea}{R}=-40E_{a}=40R$
$\ell n\left(\frac{K_{170}}{ K_{160}}\right)=\frac{Ea}{R}\left(\frac{1}{160} - \frac{1}{170}\right)$
$\ell nμ=\frac{40 R}{R}\left(\frac{10}{160 \times 170}\right)=\frac{1}{4 \times 17}=\frac{1}{68}$

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