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Rate constant for first order reaction i...

Rate constant for first order reaction is `5.78xx10^(-5)S^(-1)`. What percentage of initial reactant will react in 10 hours ?

A

`12. 5%`

B

`25%`

C

`87. 5%`

D

`75%`

Text Solution

Verified by Experts

The correct Answer is:
C
`5.78xx10^(-5)=(2.303)/(10xx3600)xx"log"(A_(0)/(A_(t)))`
` A_(0)=8ArArr=8ArArrA_(t)=(A_(0))/(8)=12.5%(A_(0))`
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