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Bad White [126]
3 years ago
14

What change would you expect on the rate of the SN2 reaction of 1-iodo-2-methylbutane with cyanide ion if the nucleophile concen

tration 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
Chemistry
1 answer:
n200080 [17]3 years ago
6 0

Answer:

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.

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Solnce55 [7]
We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+  + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
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PH= -㏒[H+]
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