After how many seconds will the concentration of the reactant in a first order reaction be halved if the rate constant is `1.155xx10^(-3)s^(-1)`?

Options

A. 600
B. 100
C. 60
D. 10

Answer

A

Explanation

Rate constant `k=1.155xx10^(-3)s^(-1)`
`k=(2.303)/(t)"log"(a)/((a-x))`
`because" "a=a,(a-x)=(a)/(2)`
`t_(1//2)=(2.303)/(k)"log"(a)/(a//2)=(2.303)/(1.155xx10^(-3))"log "2`
`or" "t_(1//2)=600s`