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In class today, we were calculating the rate law for [NOs]. 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?
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.
Matt Sanruk 2H wrote: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.
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.
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