You have studied the gas-phase oxidation of HBr by O2:
4 HBr1g2 +...

Question

You have studied the gas-phase oxidation of HBr by O2:
4 HBr1g2 + O21g2¡2 H2O1g2 + 2 Br21g2
You find the reaction to be first order with respect to HBr
and first order with respect to O2. You propose the following
mechanism:
HBr1g2 + O21g2¡ HOOBr1g2
HOOBr1g2 + HBr1g2¡2 HOBr1g2
HOBr1g2 + HBr1g2¡ H2O1g2 + Br21g2
(b) Based on the experimentally determined
rate law, which step is rate determining?

Accepted Answer

Welcome back everyone In this example, we have the following reaction Where ozone gas reacts with two moles of nitrogen dioxide gas to form oxygen gas and nitrogen or die nitrogen Penta oxide gas. We're told that this is a first order reaction with respect to both ozone and our nitrogen dioxide gas. And we're told that the following reaction mechanism is proposed. And so looking below, we have a two step mechanism we need to determine based on the rate law for our given reaction here, what is going to be the rate determining step? So we need to pick between steps one or two as our rate determining step. So the first thing to notice is that we are told that our reactant ozone and nitrogen dioxide are both first order with respect to our reaction. And that is telling us our rate law. And so we should recall that our rate law is equal to our rate constant K. Times the concentration of our reactant raised to their respective orders. However, according to our reaction we have as our reactant ozone. So we would plug in our concentration of ozone which according to the prompt is first order and then that would be racist and exploitative one here, which is then multiplied by the concentration of our second reactant given in our reaction above. In the prompt being our nitrogen dioxide. And according to the prompt. This is also a first order reactant with respect to our reaction. So we raise it to an exponent of one. And so this means we can simplify to the following rate law where we say that our rate is equal to the rate constant K. Times our concentration of ozone times the concentration of nitrogen dioxide. Because we don't really need to include those exponents of one, it is implied and understood here. And so as I stated earlier, we have a two step mechanism. So this was step one and this is step two and just so it's clear We have step one here and step two here and we need to determine which of these steps is the rate determining step. So, because our rate law includes ozone and nitrogen dioxide as our only reactant in the rate law, we want to pick the step that has these same react ints. And looking at Step one, we definitely do have the presence of ozone and we definitely do have nitrogen dioxide as well listed as react as a reactant. But looking at Step two of our mechanism given, we do not have any ozone listed, nor or we do actually have nitrogen dioxide, but we are missing ozone. So we can say missing reactant, ozone. And so we can say that because we have our ozone and our nitrogen dioxide as our rate law, reactant and ozone and nitrogen dioxide are included as reactant in step one. And we know that they're present in our rate law. We can say that therefore Step one, which we can list given in the prompt as ozone gas plus nitrogen dioxide gas produces nitrogen tri oxide gas plus oxygen gas is the rate determining step. And so this equation from Step one in our mechanism is going to be our final answer to complete this example as the rate determining step. So what's highlighted in yellow in our solution is our final answer. I hope that everything that I explained was clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.

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