Question

f the concentration of 1-iodopropane doubles and the concentration of sodium hydroxide quadruples in this SN2 reaction, how much faster is the reaction rate

Answer 1

Answer:

The reaction rate is 12times faster in SN2 reaction .

Explanation:

For SN2 reaction ,

r= k [1-iodopropane] [NaOH] -(1)

Rate depends upon reactants if we make change in concentration of reactants it will directly change the reaction.

Now concentration of 1- iodipropane is quadruple

And concentration of NaOH is tripled

$r_n_e_w$ = k 4 × [1-iodopropane] × 3 × [NaOH]

$r_n_e_w$ = 12 (k[1-iodopropane][[NaOH])

$r_n_e_w$ = 12 ( r) putting r from ( 1)

Hence , the reaction rate is 12 times faster in SN2 reaction