1
Question
A secondary alkyl halide (A) hydrolyzes with alkali (B) in aqueous medium simultaneously via SN1 and SN2 pathways with rate constants k1 and k2 respectively. From kinetic data, it was found that a plot of =−1[A]d[A]dtvs[B] is straight line with a slope equal to 2.7×10−4Lmol−1min−1 and intercept equal to 1.02×10−3. Minimum initial concentration of [A] = 0.2 M and [B], i.e., [⊖O]=0.5MThe value of overall rate constant of the hydrolysis of A (in L mol−1min−1) is:
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Solution
The correct option is D 1.29×10−3
SN1 reaction:
Rate=−d[a]dt=k1[RX]
SN2 reaction
Rtae = −d[A]dt=k2[RX][⊖OH]
Combining equations (i) and (ii),
Rate = −d[A]dt=k1[RX]+k2[RX][⊖OH]
or
−d[A]dt=k1[A]+k2[A][B]
[A][k1+k2[B]]
1A×−d[A]dt=k1+k2[B]equation of straight line:y=mx+c
∴ slope (m) =k2=2.7×10−4Lmol−1min−1
=0.27×10−3Lmol−1min−1
Intercept (c) =k1=1.02×10−3Lmol−1min−1
Overall rate constant =k1+k2
=(0.27+1.02)×10−3
=1.29×10−3Lmol−1min−1
SN1 reaction:
Rate=−d[a]dt=k1[RX]
SN2 reaction
Rtae = −d[A]dt=k2[RX][⊖OH]
Combining equations (i) and (ii),
Rate = −d[A]dt=k1[RX]+k2[RX][⊖OH]
or
−d[A]dt=k1[A]+k2[A][B]
[A][k1+k2[B]]
1A×−d[A]dt=k1+k2[B]
equation of straight line:
y=mx+c
∴ slope (m) =k2=2.7×10−4Lmol−1min−1
=0.27×10−3Lmol−1min−1
Intercept (c) =k1=1.02×10−3Lmol−1min−1
Overall rate constant =k1+k2
=(0.27+1.02)×10−3
=1.29×10−3Lmol−1min−1
∴ slope (m) =k2=2.7×10−4Lmol−1min−1
=0.27×10−3Lmol−1min−1
Intercept (c) =k1=1.02×10−3Lmol−1min−1
Overall rate constant =k1+k2
=(0.27+1.02)×10−3
=1.29×10−3Lmol−1min−1
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