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During the decomposition of H(2)O(2) to ...

During the decomposition of `H_(2)O_(2)` to gives oxygen , 48g `O_(2)` is formed per minute at a certain point of time .The rate of formation of water at this point is -

A

`0.75 "mol.min"^(-1)`

B

`1.5 "mol.min"^(-1)`

C

`2.25 "mol.min"^(-1)`

D

`3.0 "mol.min"^(-1)`

Text Solution

Verified by Experts

`2H_(2)O_(2)rarr2H_(2)O+O_(2)`
Rate `=-(1)/(2)(d[H_(2)O_(2)])/(dt)=(1)/(2)(d[H_(2)O])/(dt)=(d[O_(2)])/(dt)`
Rate of formation of oxygen = `48."min"^(-1)`
`=(48)/(32)"mol.min"^(-1)=1.5"mol.min"^(-1)`
`therefore` Rate of formation of `H_(2)O`
Rate `=(1)/(2)(d[H_(2)O])/(dt)=(d[O_(2)])/(dt)`
or, `(d[H_(2)O])/(dt)=(2d[O_(2)])/(dt)=2xx1.5=3"mol.min"^(-1)`
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