20%ofthereactioniscompletedwhichmeansthatleftamountofAis=80
Initialamountofa=100
Reactioncompletedin=10s
Henceforthecaseof50%completion
LeftamountofA=50%
InitialamountofA=100
A=Ao−kt
80=100−k×10
k=2moll−1s−1
Forthe50%completionofreaction
A=Ao−kt
50=100−2×t
2t=50
t=25s
Hencetimetakentocomplete50%ofthereactionis25s