. An acid solution of sucrose was to the exten after 67 minutes. Assuminhydrolysedg the reaction to be of the first ord late the time taken for 80% hydrolysis.
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Answer:
k=
t
1/2
0.693
=
3.0
0.693
=0.231/hr
Let initial concetration of sucrose be 1 M.
After 8 hours, the concentration will be 1−x. x represents the amount of sucrose decomposed.
k=
t
2.303
log
[A]
[A]
0
⟹0.231=
8
2.303
log(
1−x
1
)
⟹log(
1−x
1
)=0.8024
⟹
1−x
1
=6.345 ⟹x=0.842
After 8 hours, the concentration of sucrose left is 1−0.842=0.158 M.
Explanation:
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