1.

In the reaction of formation of sulphur trioxide by contact process 2SO_(2)+O_(2) harr 2SO_(3) the rate of reactionn was (d[O_(2)])/(dt)=2.5times10^(-4)mol*L^(-1)*s^(-1). The rate of reaction is terms of [SO_(2)] in mol*L^(-1)*s^(-1) will be

Answer»

`-1.25times10^(-4)`
`-2.50times10^(-4)`
`-3.75times10^(-4)`
`-5.00times10^(-4)`

Solution :From RATE law
`(1)/(2)(dSO_(2))/(DT)=-(dO_(2))/(dt)=(1)/(2)(dSO_(3))/(dt)`
`THEREFORE -(dSO_(2))/(dt)=-2times(dO_(2))/(dt)=2times2.5 times10^(-4)=-5times10^(-4)MOL*L^(-1)*s^(-1)`


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