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.  

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm