The rate equation for the second-order reaction of an enzyme (e) and its substrate (s) is vo = keat le][s] near the where it is the second-order rate constant, which can be used to compare catalytic efficiencies. enzymes with a diffusion-controlled limit are said to have achieved catalytic perfection. what is the diffusion-controlled limit in an aqueous solution? 10-4 to 10-8 m-s-1 107 to 10% m s 10 to 10-sm 109 to 1010 m.5-1 o 10-7 to 10-'m-'s-1 10% to 10º m/s o 10% to 10% m-1-1

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

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i think its 5049.

i hope this !