Week 9 Wednesday lecture

[Megan Cao 1I]

In class today, we were calculating the rate law for [NO_s]. the chemical equation was NO₂ + CO --> NO + CO₂ (all are gases). for the overall rate law we got K₁[NO₂]². What happened to the CO? since we're supposed to include all reactants in the rate law, is it the zeroth order?

[Matt Sanruk 2H]

Since it is not present in the slowest step we can assume that it is indeed zero order and does not affect the rate

[Shail Avasthi 2C]

CO is not in the rate law because its concentration does not affect the rate of the reaction (i.e. it is zero order). This implies that CO is not involved in the slowest step of this multi-step reaction. It does not mean that CO is not involved in the reaction at all. Just that it's not significant to how fast the reaction proceeds.

[BCaballero_4F]

CO is zero order and because it is not in the slow step e assume its concentration does not have an affect overall

[JohannaPerezH2F]

yes it would be in the zeroth order

[Matt Sanruk 2H]

But a solvent is different right?

[Jialun Chen 4F]

Since it is not present in the slowest step we can assume that it is indeed zero order and does not affect the rate

I agree with Matt's rationale, the concentration of a zero-order has no effect on reaction.

[John Liang 2I]

when we separated it out into fast and slow steps, we ignore the fast step reactants and products because these do not control the rate. an example used in class is when friends bake brownies with each person on a specific task, the rate of the production of brownies is dependent on the one slow friend, rather than the 5 other fast friends. hope this helps.