a) What will be the initial rate of reaction if rate constant is `10^(-3)s^(-1)` at concentration of 0.2 mol `L^(-1)`? How much of the reactant will be converted into product in 200 minutes? Assume reaction to be of first order.
b) The half life of 1st order reaction `A to B` is 600s. What percentage of A remains after 30 minutes?

Explanation

a) Calculation of initial rate of reaction
for the first order reaction,
Reaction rate=`k[A]=(10^(-3)min^(-1)) xx (0.2 mol L^(-1))`
`=2.0 xx 10^(-4)mol L^(-1)min^(-1)`
Calculation of amount of reactant changing into products.
For the first order reaction,
`t=(2.303)/k log ([A]_(0))/([A]), (200 min) = (2.303)/(10^(-3)min^(-1)) log ([A]_(0))/([A])`
`log([A])_(0)/[A] = (200 min xx 10^(-3) min^(-1))/(2.303) = 0.0868`
`[A]_(0)/[A] = "Antilog" 0.0868=1.221` or `[A]=([A]_(0))/(1.221) = (0.2 mol L^(-1))/(1.221) = 0.164 mol L^(-1)`
`therefore` Amount of rectant changing into product =`0.2-0.164=0.036 mol L^(-1)`
b) Calculation of the percentage of A left after 30 minutes.
For the first order reaction,
`t=(2.303)/k log ([A]_(0))/([A]), (30 xx 60s) = (2.303)/(0.001155 s^(-1)) log ([A]_(0))/([A])`
`log ([A]_(0))/[A] = (30 xx 60s xx 0.001155 s^(-1))/(2.303) = 0.9027`
`([A]_(0))/([A]) = "Antilog" 0.9027 = 7.9928, [A] = ([A]_(0))/(7.9928) = 100/(7.9928) = 12.51%`