Answer
See the answer below.
Work Step by Step
a) Reacted concentration: x
$K=[dim]/[mon]^2$
$3.2\times10^4=x^2/(5.4\times10^{-4}-2x)$
$x^2+6.4\times10^4x-17.28=0$
$x=2.70\times10^{-4}\ M$
$f=(1-2x/5.4\times10^{-4})\times100\%\approx100\%$
b) Since the reaction is exothermic, as the temperature increases the equilibrium shifts towards the monomer.
