A first order gas reaction has `k = 1.5 xx 10^(-6) s^(-1)` at `200^(@)C`. If the reaction is allowed to run for `10 h`, what have changed in the Product? What is the half-life of this reaction?

Answer

`t_(1//2) = 128.33 hr`

Explanation

For a first order reaction,
`k = (2.303)/(t)log.(A_(0))/(A_(t))`
`= (2.303)/(t)log.(A_(0))/((A_(0) - x))`
or `1.5 xx 10^(-6) = (2.303)/(3600 xx 10s)log.(100)/((100-x))`
or `x = 5.2`
Thus, the initial concentration changed into Product is `52%`.
Half life `= (0.693)/(k) = (0.693)/(1.5 xx 10^(-6))`
`= 462000 s`
`= 128.33 hr`