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Decomposition of H(2)O(5) follows a fi...

Decomposition of `H_(2)O_(5)` follows a first order reaction . In fifty minutes , the concentration of `H_(2)O_(2)` decreases from 0.5 to 0.125 (M) in one such decomposition when the concentration of `H_(2)O_(2)` reaches 0.05 (M) , the rate of formation of `O_(2)` will be -

A

`6.93xx10^(-2)"mol.min"^(-1)`

B

`6.93xx10^(-4)"mol.L"^(-1)"min"^(-1)`

C

`2.66"L.min"^(-1)` at STP

D

`1.34xx10^(-2)"mol.min"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
The decomposition reaction of `H_(2)O_(2)` is -
`H_(2)O_(2)(l)rarrH_(2)O(l)+(1)/(2)O_(2)(g)`
`therefore " Rate constant", k=(2.303)/(t)log.([A]_(0))/([A])=(2.303)/(50)log.(0.5)/(0.125)`
`= 0.0277"min"^(-1)`
Now, `-(d[H_(2)O_(2)])/(dt)=2=(d[O_(2)])/(dt)=k[H_(2)O_(2)]`
or, `2(d[O_(2)])/(dt)=0.0277xx0.05`
`therefore(d[O_(2)])/(dt)=6.93xx10^(-4)"mol.L"^(-1)."min"^(-1)`
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