Starting with one mole of a compound `A`, it is found that the reaction is `3//4` completed in `1 hr`. Calculate the rate constant if the reaction is of
a. First order , b. Second order

Explanation

`a = 1 "mole", x = 3//4 "mole", t = 1 hr`
a. First order: `k = (2.303)/(1) log.(1)/(1-(3)/(4))`
`= (2.303)/(1) log4 = 1.386 hr^(-1)`
b. Second order: `k = (1)/(t) xx (x)/(a(a-x))`
`= (1)/(1) xx (3//4)/(1 xx (1-3//4)) = 3`
`k = 3 L mol^(-1) hr^(-1)`