 Solution

The initial ozone concentration is $(3\times 10^{-6})[M]=3.60\times 10^{12}\mathrm{molecules}\mathrm{c}{\mathrm{m}}^{-3}$. Over fifty days with twelve active hours per day, the required combined daytime rate isThe mixed chlorine-bromine ozone-loss cycle also consumes two ozone molecules per radical-pair reaction, so $L_{\mathrm{Br}}=2k_6[\mathrm{ClO}][\mathrm{BrO}]$. Subtracting the dimer contribution yieldsThe same result follows by assigning the dimer $1.897$ ppm of the fifty-day loss and the mixed cycle the remaining $1.103$ ppm. This uses the same daylight fraction for both cycles and retains the earlier fixed-ClO approximation. It estimates a sunlit BrO abundance, not the total inventory of bromine in all reservoirs.