1.

The rate of the reaction 2SO_(2) +O_(2) rarr 2SO_(3) may be expressed as : -(d[O_(2)])/(dt)= 2.5 xx10^(-4) "mol L"^(-1) "sec"^(-1) The rate of reaction when expressed in terms of SO_(3) will be :

Answer»

`-5.0xx10^(-4) "mol L"^(-1) "SEC"^(-1)`
`-1.25xx10^(-4) "mol L"^(-1) "sec"^(-1)`
`3.75xx10^(-4) "mol L"^(-1) "sec"^(-1) `
`5.0xx10^(-4) "mol L"^(4) "sec"^(-1)`

Solution :(D) Rate `= -(1)/(2) (d[SO_(2)])/(DT) =-(d[O_(2)])/(dt)`
`(d[SO_(2)])/(dt)=2 (d[O_(2)])/(dt)`
`=2xx2.5 xx10^(-4) `
`= 5.0 xx10^(-4)mol L^(-1) sec^(-I)`


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