Decomposition of a gas is of first order. It takes 80 minutes for 80% of the gas to be decomposed when its initial concentration is `8 xx 10^(-3)` mole/litre. Calculate the specific reaction rate.

Answer

`0.02012 min""^(-1)`

Explanation

`t=( 2.303 )/(k ) log "" (a)/((a-x)) ` where ` a= 8xx 10^(-3) mol L^(-1)`
` x= ( 8 xx 10 ^(-3) xx 80 )/( 100 ) =6.4 xx 10^(-3)`
` ( a- x) = (8 xx 10^(-3) - 6.4 xx 10^(-3))`
` = 1.6 xx 10^(-3) mol L^(-1)`
Substituting the values,
`80= ( 2.303 )/( k ) log"" ( 8 xx 10^(-3))/( 1.6 xx 10^(-3)) = ( 2.303 ) /(k ) log 5= (2.303 )/(k ) xx 0.6989 `
` or k= ( 2.303 xx 0.693 )/( 80 min ) = 0.02012 min""^(-1)`