Question

the rate constant of a first order reaction A --> B is 1.155* 10^(-3)s^(-1).the percent of A remaining after 1 hour will be antilog (1.805)=63.826 ​

Answer 1

Answer:

93%

Explanation:

For a first order Reaction,

⇒ $k= \frac{2.303}{t}$× $log[ \frac{R_0}{R} ]$

⇒ $log[ \frac{R_0}{R} ]$ = $\frac{kt}{2.303}$

⇒$log[ \frac{R_0}{R} ]$ = $\frac{1.155 *10^{-3} *60}{2.303}$

⇒$log[ \frac{R_0}{R} ]$ = 0.03

⇒$\frac{R_{0} }{R}$= 1.0715

⇒$\frac{R}{R_{0} }$ = 0.93

∴ % age= 93%

QED