The given question is incomplete. The complete question is :
Consider the following SN2 reaction,
1-chloro-3-methylbutane + [tex]NaN_3\rightarrow products[/tex]
Assuming no other changes, what is the effect on the rate, if the concentration of 1-chloro-3-methylbutane is doubled? It would increase four times. No effect. It would double the rate. It would reduce by half. It would triple the rate.
Answer: It would double the rate.
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
As the reaction follows [tex]SN_2[/tex] mechanism, the rate depends on the concentration of both 1-chloro-3-methylbutane and sodium azide.
[tex]rate=k[1-chloro-3-methylbutane]^1[NaN_3]^1[/tex]
k= rate constant
As concentration of 1-chloro-3-methylbutane is doubled,
[tex]rate'=k[2\times 1-chloro-3-methylbutane]^1[NaN_3]^1[/tex]
[tex]rate'=2^1\times k[ 1-chloro-3-methylbutane]^1[NaN_3]^1[/tex]
[tex]rate'=2\times rate[/tex]
Thus it would double the rate.
Doubling the initial concentration of 1-chloro-3-methylbutane will double the rate of reaction.
An SN2 reaction is a synchronous reaction. The nucleophile enters as the leaving group departs in a single transition state. The rate of reactions depends on the concentration of the alkyl halide and the nucleophile.
If we represent the nucleophile in this reaction as Nu-, the rate of reaction is written as;
Rate =k [1-chloro-3-methylbutane] [Nu-]
When the concentration of 1-chloro-3-methylbutane is doubled, we have;
Rate' = k'[1-chloro-3-methylbutane]^2 [Nu-]
We can see from the equation above that after doubling the initial concentration of 1-chloro-3-methylbutane will we succeed in doubling the rate of reaction.
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