A first order reaction has a specific reaction rate of `10^(-2) S^(-1)`. How much time will it take for 20 g of the reactant to reduce to 5 g?

Options

A. 238.6s
B. 138.6s
C. 346.5s
D. 693.0s

Answer

B

Explanation

for a first order reaction
rate constant `(K) =(2303)/(t) log (a) /(a-x)`
where a initial concentration
`a-x=`concentration after time 't'
.t= time in sec
given a=20 g,a-x =5g.k =`10^(-2)`
`therefore t=(2303)/(10^(-2))log (20)/(5) `
`=1386 s`
Alternatively ,
Half - life for the first order reaction
`(t_(1//2))/(2)=(0.693)/(10^(-2))=69.3 s`
two half -lives are required for the reduction of 20 g of reactant into 5g
`20 g overset( t_(1//2)) to 10g overset( t_(1//2))to 5g.`
`therefore "total time "=2t_(1//2)=2xx69.3=138.6s`