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 contant `K=1.55xx10^(-3)s^(-1)`
`k=(2.303)/(k)"log"(a)/(a//2)`
` =(2.303)/(1.55xx10^(-3)) log2`
`=(2.303)/(.1.5xx10^(-3))xx0.3010`
`=(0.693xx10^(3))/(1.155)`
or `t_(1//2)=(0.693)/(k)=(0.693)/(1.55xx10^(-3))`
`=600s`