WizardOfBoz

Kind of, depending upon definitions. If "V" is the viral load in a certain volume, and we add a specific volume of sanitizing solution of concentration S, then assuming first order kinetics you get

rate of reduction in V with respect to time equals k times V times S, that is:

dV/dt = rate = k * V * S

So the overall rate of virus destruction is proportional to the amount of virus present. I'm guessing that this works for bleach, and possibly for quaternary ammonium compounds (dilute solutions that react with the virus particles as describe above), but not for ethanol and isopropanol (strong, 70% solutions that work via bulk physical chemistry, probably by "dissolving" the virus lipid coat). This seems consistent with the data reported (see the NEJM article cited in my next post).

But generally we're interested in fractional rate of virus change. For this, the reporting is most usefully done in log terms. This is easily done:

dV/dt /V = d log(V)/dat = log rate = k * S

Assuming S is much larger than V and only reacts with virus, S remains pretty constant. If you want to get fancier, you can calculate a half-life. Solving differential equation, above yields

log V/Vo = -k*S*t

log(0.5) = -k * S * thalf

thalf = ln(2)/(kS)

So, doubling the sterilizing solution concentration S should cut the half-life of Virus particles in half. Again, the equations above that use the concetnration S of sanitizer probably work for dilute solutions (bleach, quaternary ammonium compounds). I suspect that it breaks down for stuff that requires strong solutions to be effective at all, eg. ethanol and isopropanol. For the latter, use a 70% solution.