The half-life period of a first order reaction is 35 minutes. What fraction of the reactant remains after 75 minutes?

Options

A. 4.415
B. 0.226
C. 5.263
D. 0.155

Answer

B

Explanation

`k=(0.693)/(t_(1//2)=(0.693)/(35)` minute`s^(-1)`
`k=(2.303)/(t)log(a)/(a-x)`
here, (a - x) is the undecomposed reactant. We need to calculate the fraction of reactant left i.e., `(a-x)/(a)`
`(0.693)/(35)=(2.303)/(75)log(a)/(a-x)implieslog(a)/(a-x)=(0.693xx75)/(35xx2.303)=0.645`
Taking antilog,`(a)/(a-x)`=4.415
`(a-x)/(a)=(1)/(4.415)=0.226`