Question

In a I order reaction `A rarr` products, the concentration of the reactant decrease to `6.25%` of its initial value in `80` minutes. What is (i) the rate constant and (ii) the rate of the reaction, `100` minutes after the start, if the initial concentration is `0.2 "mole"//"litre"` ?
A. `2.17 xx 10^(-2) "min"^(-1), 3.47 xx 10^(-4) "mol.litre"^(-1) "min"^(-1)`
B. `3.465 xx 10^(-2) "min"^(-1), 2.166 xx 10^(-4) "mol.litre"^(-1) "min"^(-1)`
C. `3.465 xx 10^(-3) "min"^(-1), 2.17 xx 10^(-3) "mol.litre"^(-1) "min"^(-1)`
D. ( d) `2.166 xx 10^(-3) "min"^(-1), 2.667 xx 10^(-4) "mol.litre"^(-1) "min"^(-1)`

Answer 1

Correct Answer - B
`K = (2.303)/(80)"log"((100)/(6.25)) = 3.465 xx 10^(-2)mm^(-1)`
Then `3.465 xx 10^(-2) = (2.303)/(100)"log"((0.2)/(a_(t)))`
`a_(t) = 0.00625`
Rate `= K xx [a_(t)]`
`= 0.00625 xx 3.465 xx 10^(-2)`
`= 2.166 xx 10^(-4) sec^(-1)`