Answer
$9.5\times10^{-5}L/m•s$
Work Step by Step
We know that
log $k_{2}$-log $k_{1}$= $\frac{E_{a}}{2.303R}[\frac{1}{T_{1}}-\frac{1}{T_{2}}]$
log $k_{2}$=log $k_{1}+\frac{E_{a}}{2.303R}[\frac{1}{T_{1}}-\frac{1}{T_{2}}]$
= log$(3.52\times10^{-7})+ \frac{186000J/mol}{2.303\times8.314JK^{-1}mol^{-1}}[\frac{1}{555K}-\frac{1}{645K}]$
log $k_{2}$= -6.45+2.43= -4.02
$k_{2}$ = $9.5\times10^{-5}L/m•s$