# The yields of reactions with large driving energies are insensitive to small variations in driving energy

Large energy differentials can provide a margin of safety against errors in modeling. For example, in a mechanosynthetic system, positioning a tip against a workpiece will commonly create a two-state system, S_{0} = unreacted, S_{1} = reacted. If the change in the free energy of the reactants (from pre- to post-reaction) is Δ*E*_{01}*,* then at equilibrium the probability ratio *R*_{01} = P(S_{0})/P(S_{1}) = exp(Δ*E*_{01}/*kT*), and the probability of failure = P(S_{0}) = *R*_{01}/(1 + *R*_{01}). If a design calculation estimates Δ*E*_{01} = −4*kT,* then an adverse error of +3*kT* would drop the probability of success from a calculated 0.98 to an actual 0.73. If the calculated value is a more substantial −14*kT,* however, an adverse error of +3*kT* drops the probability merely from 0.99999917 to 0.999983. Reactions that produce a net gain of one covalent bond (for example, radical coupling) commonly reduce free energy by more than 100*kT*_{300}. Reactions that break weak bonds in order to form strong ones (for example, abstraction of hydrogen from tin to carbon) commonly reduce free energy by more than 30*kT*_{300}.