Decompsition of `H_(2)O_(2)` follows a frist order reactions. In 50 min the concentrations 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 fromation of `O_(2)` will be

Options

A. `6.93xx10^(-4)" mol m in"^(-1)`
B. `2.66" L m in"^(-1)` at STP
C. `1.34xx10^(-2)" mol m in"^(-1)`
D. `6.93xx10^(-2)" mol m in"^(-1)`

Answer

A

Explanation

In fifty minutes the concentration of `H_(2)O_(2)` decreases from 0.5 to 0.125 M. It means two half lives must have passed
`rArr 2t_(1//2)=50` minutes
`t_(1//2) =25" minutes "therefore k=((0.693)/(25))" m in"^(-1)`
Also `(-d[H_(2)O_(2)])/(dt) =k [H_(2)O_(2)]=(0.693)/(25)xx(0.05)" mol m in"^(-1)`
As per reaction
`2H_(2)O_(2) rarr 2H_(2)O +O_(2)`
`(d[O_(2)])/(dt) =-(1)/(2) ((d[H_(2)O_(2)])/(dt))=(1)/(2)xx(0.693)/(25)xx0.05" mol m in"^(-1)=6.93xx10^(-4)" mol m in"^(-1)`