1.

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%`

Answer» Correct Answer - 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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