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

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

= k 4 × [1-iodopropane] × 3 × [NaOH]

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

= 12 ( r) putting r from ( 1)

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

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

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