The gas-phase decomposition of ozone is thought to occur
by the f...

Question

The gas-phase decomposition of ozone is thought to occur
by the following two-step mechanism.
Step 1: O31g2ΔO21g2 + O1g2 (fast)
Step 2: O1g2 + O31g2¡2 O21g2 (slow)
(d) If instead the reaction
occurred in a single step, would the rate law change? If so,
what would it be?

Accepted Answer

Hello everyone today. We are being asked to consider a formal reaction with a two step mechanism. So we have step one and we have step to note that step one is the slow step and this is going to come into importance later on. We are then asked if the rate law would change if the reaction occurred in a single step. If yes, what would the new rate law be? So the first thing you wanna do is you want to account for what this overall reaction would look like. So we're gonna look at what the overall reaction look like. First. By doing that, we take our step one, which is that two moles of N. 02 that form one, N. 03 plus one nitric oxide. And we're going to go and add it to the second step which is N. 03 plus carbon monoxide to form N. 02 and C. 02, we are then going to get rid of any molecules on opposite sides, the same molecules and opposite sides are going to subtract from each other and the same molecules on the same side will add. So we can note that for N. 02, we have one on the left here for the first reaction. The first step and then we have one on the right for the second reaction. So we're then going to be left with only one N. 02 at the on the first step on the left, we also note that no three is present on opposite sides of both step one and two. And so since there's only one of each, they both go away and thus we are left with the overall reaction. So the overall it's going to be no two plus C. O. Or carbon monoxide to form in oh and carbon dioxide. And so the rate law of the original. So we'll say rate law of the original reaction is going to be based off of the slow step. And we already set the slow step. Is that two N. 02 molecules. And so the rate is going to be equal to some constant K. Multiplied by our two molecules being used in 02. And since there's two, we're going to use an exponent of two. So this is the rate law of the original one. So if there was only one step we would then consider the overall reaction. And so the overall reaction uses N. 02 as well as ceo. So that rate law. So rate law for one step would be that the rate is equal to some constant K. Times N. 02 times C. O. And that is going to be our final answer. I hope this helps. And until next time

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism. Step 1: O31g2ΔO21g2 + O1g2 (fast) Step 2: O1g2 + O31g2¡2 O21g2 (slow) (d) If instead the reaction occurred in a single step, would the rate law change? If so, what would it be?

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