A first order reaction has `k=1.5xx10^(-6)` per second at `240^(@)C`. If the reaction is allowed to run for 10 hours, what percentage of initial concentration would have changed to products ? What is the half-life period of this reaction ?

Answer

`5.2%,128.3" hrs"`

Explanation

`1.5xx10^(-6)s^(-1)=(2.303)/(10xx60xx60s)log""(a)/(a-x)" or "log""(a)/(a-x)=0.0234`
or `(a)/(a-x)=1.055" or "a=1.055a-1.055x" or "1.055a" or "(x)/(a)=(0.055)/(1.055)=0.052=5.2%`
`t_(1//2)=(0.693)/(1.5xx10^(-6)s^(-1))=4.62xx10^(5)s=(4.62xx10^(5))/(60xx60)" hr "=128.3" hrs"`