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?

Options

A. `5.3%`
B. `94.8%`
C. `6.7%`
D. `93.3%`

Answer

A

Explanation

Find fraction remaining `(a-x//a)` and then fraction reaction `(x//a)` . According to the integrated rate law for the first order reaction
`k=(2.303)/(t)log(a)/(a-x)`
Or, log `(a-x)/(a)=(kt)/(2.303)`
`=-((1.5xx10^(-6)s^(-1))(10xx60xx60s))/(2.303)`
`=-0.02344`
Taking the antilog we get
`(a-x)/(a)=0.9473`
Thus, precentage of reactant remaining
`=(a-x//a)xx100%`
`=94.73%`
Hence, precentage of initial concentration converted into product will be `(100-94.73)%=5.27%`
Alternatively, take `a=100` and find `x` . Then
`k=(2.303)/(t)log(a)/(a-x)`
`1.5xx10^(-6)s^(-1)=(2.303)/(3.6xx10^(4)s)log(100)/(100-x)`
or, use the following expression
`k=(2.303)/(t_(f))log(1)/(1-f)`
and calculate `'f'` the fraction of reaction completed.