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

Answer

B

Explanation

For first order kinetics
`k=(2.303)/(t)log.([R_(0)])/([R_(t)])`
`=(2.303)/(50)log[(0.5)/(0.125)]`
`0.0277 "min"^(-1)` Rate of decomposition of `H_(2)O_(2)=k[H_(2)O_(2)]`
`=0.0277[0.05]`
or `(-d[H_(2)O_(5)])/(dt)=1.386xx10^(-3) M"min"^(-1)`
`2H_(2)O_(2)rarr2H_(2)O+O_(2)`
`(+d[O_(2)])/(dt)=(-1)/(2)(d[H_(2)O_(2)])/(dt)`
`=(1)/(2)xx1.386xx10^(-4)`
`=6.93xx10^(-4) M"min"^(-1)`