Consider this three-step mechanism for a reaction:
Cl2( g) Δk1k22...

Question

Consider this three-step mechanism for a reaction:
Cl2( g) Δk1k22 Cl( g) Fast
Cl( g) + CHCl3( g) ¡k3 HCl( g) + CCl3( g) Slow
Cl( g) + CCl3( g) ¡k4 CCl4( g) Fast
a. What is the overall reaction?

Accepted Answer

Hello everyone today. We are being asked to write an overall equation for the overall reaction using the above mechanism. So we have step one, step two and step three. So the first thing you want to do is you want to write these steps out? So we'll do just that will say we have to gas reacting to form two most of bromine bromine gas. We also have that bro mean gas forming with methane in the gas form to form hbr which is a strong acid in the gas form and Methane or methyl group in the gas form. And then we have that CH three reacting with our roaming from earlier To form this CH three be our product. And so we want to remember a couple of rules. The first one is that the same species? On the same side they're going to add and the same species this is going to be the reverse on different sides. So different sides, they are going to go ahead and subtract using that rule we can go ahead and proceed, we see that we have a we have one bro man on the right on the first equation and we have one brunette on the left for the second equation. Therefore they're going to cancel out. We also see in a second equation we have a CH three on the right of the arrow and we also have a CH three on the left of the arrow. Of the subsequent reaction? We were left with one bro mean in the first equation since we had to and only got rid of one. However, notice how the third reaction also has a bro mean, so we can go ahead and get rid of that second bro mean, and thus we are left with the remaining equation that we have our bro mean, gas, BR two gas plus our methane CH four, also in the gas, and that is going to go ahead and form CH three B are in our gas form as well as H. B. R. In the gas form as well. I hope this helped, and until next time.

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