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Nitrogen(II) oxide, NO, reacts with hydrogen, ${H}_{2},$ according to the following equation:$2 {NO}+2 {H}_{2} \longrightarrow {N}_{2}+2 {H}_{2} {O}$What would the rate law be if the mechanism for this reaction were:$2 {NO}+{H}_{2} \longrightarrow {N}_{2}+{H}_{2} {O}_{2}(\text { slow) }$${H}_{2} {O}_{2}+{H}_{2} \longrightarrow 2 {H}_{2} {O}(\text { fast) }$
$\mathbf{r}=\mathrm{k}[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right]$
Chemistry 102
Chapter 12
Kinetics
Drexel University
Brown University
University of Toronto
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level on, says Ricky. And today we're going to be writing the rate law for a two step mechanism. So you have to and, oh, plus age to makes and two plus H 202 when h 202 plus age too makes to H 20 And this step is fast. Step is slow. So when we're dealing about rate laws for multi step reactions, we only care about slow step, which is called the right determining step. And so that means if we were to write out the rate law we have our, um our is equal to K to the and oh squared two h two. So I hope this video is helpful, you know?
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