Tetrachloromethane is produced from the substitution reaction between methane and chlorine gas.If the rate of formation of tetrachloromethane is 0.05 mol dm-³ min-¹,what is the rate of disappearance of chlorine gas?
A.0.0125 mol dm-³ min-¹
B.0.0500 mol dm-³ min-¹
C.0.100 mol dm-³ min-¹
D.0.200 mol dm-³ min-¹
Answers
Multiple substitution in the methane and chlorine reaction
As the amount of trichloromethane builds up, then you will get these steps giving tetrachloromethane:
This is why you will always get a mixture of products whatever the reaction proportions of methane and chlorine you use. The whole process is simply governed by chance. Having produced some chloromethane there is no way that you can prevent it from being hit by chlorine radicals, and similarly for dichloromethane and trichloromethane.
Trying to produce mainly one product
If you wanted tetrachloromethane, you could of course get it by using a large excess of chlorine, so that eventually all the hydrogens would be replaced.
If you wanted mainly chloromethane, you could favour this by using a huge excess of methane so that the chances were always greater of a chlorine radical hitting a methane rather than anything else - but even so, you would still get some mixture of products.
There is no obvious way of getting mainly dichloromethane or trichloromethane.
Answer:
option C is correct
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