What change would you expect on the rate of the SN2 reaction of 1-iodo-2-methylbutane with cyanide ion if the nucleophile concentration is halved and the alkyl halide concentration is doubled? a. divided by 4
b. halved
c. no effect
d. increased 2x
e. increased 3x
f. increased 6x
g. increased 9x

no effect

Explanation:

In an SN2 reaction, the mechanism is bi-molecular and first order in both the alkyl halide and the nucleophile.

Hence the rate of reaction is;

Rate = k [1-iodo-2-methylbutane] [cyanide ion]

Given that it is a bimolecular reaction, if we double the concentration of 1-iodo-2-methylbutane and the concentration of the cyanide ion is halved, the rate of reaction remains the same.

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

5.6⋅10−^19  j

do you need the explaination?