Im confused, what is it that we would actually have to do in a question since the ib state that we dont have to propose a reaction mechanism in the current syllabus, could you please help? :)
First exams 2025: Evaluate proposed reaction mechanisms and recognise intermediates. Distinguish between intermediates and transition states, and recognise both in energy profiles of reactions. Include examples where the rate-determining step is not the first step. Proposed reaction mechanisms must be consistent with kinetic and stoichiometric data.
They are still there they don't cancel out, but they (K1, K-1 and K2) can be simplified in to a new constant (seems confusing cause he uses K2 again). That doesn't matter though as the orders of reaction matches the ones of the expermental one
@@MSJChem oh I get it, if the slow step is the first step, I can just look at that to get the rate. But if the slow step is the 2nd step, I can just look at the overral equation and get the rate.
Why is the rate expression [NO2]2 if you ignore the intermediates? wouldn't it be first order?
You have to include the O2 in step 1.
@@MSJChem For the reaction at 5:37?
@@LloydJackson-bx1zctwo reactant particles is bimolecular
@@MSJChem oh right so you don't remove it even if it is an intermediate?
NO3 is the intermediate.
for the question before the last one, why can step 1's reverse rate have NO3 it in but step 2's rate cannot?
Im confused, what is it that we would actually have to do in a question since the ib state that we dont have to propose a reaction mechanism in the current syllabus, could you please help? :)
wait really??? is it for the new '25 exam???
First exams 2025:
Evaluate proposed reaction mechanisms and recognise intermediates.
Distinguish between intermediates and transition states, and recognise both in energy profiles of reactions.
Include examples where the rate-determining step is not the first step.
Proposed reaction mechanisms must be consistent with kinetic and stoichiometric data.
@@mskm_vlogs_sai idk I was 2024 I’m glad it’s all in the past, but good luck and listen to king MSJchem
Does k1 and k-1 cancel out because they are the same in equilibrium?
They are still there they don't cancel out, but they (K1, K-1 and K2) can be simplified in to a new constant (seems confusing cause he uses K2 again). That doesn't matter though as the orders of reaction matches the ones of the expermental one
@@MSJChem Isn't k-1 the reciprocal of k1? Or are they the same exact value?
can I just look at the slow step and its coefficients to find the rate?
Only if it's the first step.
@@MSJChem oh I get it, if the slow step is the first step, I can just look at that to get the rate. But if the slow step is the 2nd step, I can just look at the overral equation and get the rate.
If the slow step is the second step you need to look at the preceding step.
So does that mean we can tell the order with respect to something based on if the slow step is uni or bimolecular?
Nice explanation
You had me until the equilibrium and elimination of intermediate
Ty :)
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