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.  

hey there!

boiling point is the temperature at which the vapor pressure of the liquid equals atmospheric pressure. the vapor pressure of solvent molecules is decreased when a solute is added, so a higher temperature is required to increase the number of solvent molecules in the gas phase above the liquid.

at the freezing point, the vapor pressures of the solid and liquid are equal; a lower temperature is needed to reduce the number of solvent particles above the liquid.

hope this you

tobey