Rate of formation `SO_(3)` in this reaction is `1.6xx10^(-3)kg//min`
`2SO_(2)+O_(2)to2SO_(3)` then rate at which ` SO_(2)` reacts is :-

Options

A. `1.6xx10^(-3) kg//mi n^(-1)`
B. `8xx10^(-4) kg//mi n^(-1)`
C. `3.2xx10^(-3)kg mi n^(-1)`
D. `1.28xx10^(-3)kg mi n^(-1)`

Answer

D

Explanation

`(dSO_(3))/(dt)=1.6xx10^(-3) Kg//min`
`=(1.6xx10^(-3)1000)/(80)mol//min`
`(dSO_(2))/(2dt)-(-dO_(2))/(dt)=(dSO_(3))/(2dt)`
`(-dSO_(2))/(dt)=(1.6)/(80)mol//min`
`(1.6)/(80)xx64xx10^(-3)kg//mol`
`=1.28xx10^(-3)mol//min`