Consider the following SN2 reaction, 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.

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.

Learn more: link

The given question is incomplete. The complete question is :

Consider the following SN2 reaction,

1-chloro-3-methylbutane +

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.

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 mechanism, the rate depends on the concentration of both 1-chloro-3-methylbutane and sodium azide.

k= rate constant

As concentration of 1-chloro-3-methylbutane is doubled,

Thus it would double the rate.

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"