woahh. manifesting this even tho i have less than 2 weeks to bring my gpa up

Hi bro From India

@@hershnotfound hi bro from srilanka

❤️

SOSSSSSSS I AM CRACKED HHHAKLJJJSAKAOPKSAJJSJSJAJJAJJAJAJAAJJAA

Yeah don't understand it either. When I calculate the slope I get -0.00389. Not sure what I'm doing wrong?

Hello! Good question! Based on Problem 4.3.4., Step 2 was given as the "slow" or rate determining step, so we wrote down the rate of that step as the overall reaction rate. It just so happens that HOCl is the only intermediate that appears in the reaction rate expression, so we did a steady-state approximation to solve for an expression for HOCl. We could have also written a steady-state approximation expression for the other intermediates, but that for HOCl is the most useful for us in the problem. Alternatively for a more general approach... if we don't know which is the slow step, we can write down the rate of the reaction in more general terms, like the concentration change of product with time, either Cl- or OI- (or we can write it in terms of reactant too, but add a negative sign). If we use Cl-, we get: (eq. 1) Rate = d[Cl-]/dt= k2[I-][HOCl] (same if we knew that step 2 was the slowest step!) OR, if we use OI-, we get: (eq. 2) Rate= d[OI-]/dt=k3[HOI][OH-] - k-3[OI-] (note that we're assuming H2O is in very large excess, so its concentration is considered to be more or less constant) If we use the rate of OI-, we'll have to write the steady state approximation for HOI since it appears in the expression. Interestingly, if we plug an expression for HOI into eq. 2, it just simplifies to eq. 1. So using the change in concentration of either product with time as the overall rate of reaction should yield the same results. I hope this helps! :)

Oops, my mistake! Yes, it should be dt! I think mentally I got ahead of myself and had the integrated form in mind :) Thanks very much for pointing this out :)