Consider the reaction of 2-chloro-2-methylpentane with sodium iodide. Assuming no other changes, how would it affect the rate if one simultaneously doubled the concentration of 2-chloro-2-methylpentane and sodium iodide? Group of answer choices

The given question is incomplete, the complete question is:

Consider the reaction of 2-chloro-2-methylpentane with sodium iodide. Assuming no other changes, how would it affect the rate if one simultaneously doubled the concentration of 2- chloro-2-methylpentane and sodium iodide? + NaCI + Nal CI A) no effect B) it would double the rate C) it would triple the rate D) it would quadruple the rate E) it would increase the rate 5 times

The correct answer is option B, that is, it would double the rate.

Explanation:

Based on the given reaction, the reactant, that is, 2-chloro-2-methylpentane is a sterically crowded molecule, therefore, the formation of the product will take place by SN₁ reaction mechanism. The formation of carbocation takes place in the initial step and is considered as the rate-determining step.  

The rate equation is R = K₁ [A]ⁿ

Here A is 2-chloro-2-methylpentane, therefore, if the concentration of A increases, the rate of the reaction will get double. Now the rate becomes, R = K₁ [A]²  

In the given case, NaI or sodium iodide is a nucleophile, and the concentration of nucleophile is not present in the rate equation, it is a unimolecular reaction. So, option 1 cannot be correct.  

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current