Decomposition of `H_(2)O_(2)` follows a first order reactions. In 50 min 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

Options

A. `366 L min^(-1)` at STP
B. `1.34 xx 10^(-2) mol min^(-1)`
C. `6.96 xx 10^(-2) mol min^(-1)`
D. `6.93 xx 10^(-4) mol min^(-1)`

Answer

D

Explanation

`H_(2)O_(2)(aq) to H_(2)O(aq)+(1)/(2)O_(2)(g)`
For a first order reaction
`k=(2.303)/(t)log""(a)/((a-x))`
Given a = 0.5, (a-x) = 0.125, t = 50 min
`:. k=(2.303)/(50) log""(0.5)/(0.125), =2.78 xx 10^(-2) min^(-1)`
`r=k[H_(2)O_(2)]=2.78 xx 10^(-2) xx 0.05`
`=1.386 xx 10^(-3) mol min^(-1)`
Now
`-(d[H_(2)O_(2)])/(dt)=(d[H_(2)O])/(dt)=(2d[O_(2)])/(dt)`
`:. (d[O_(2)])/(dt)=(1)/(2) xx (d[H_(2)O_(2)])/(dt)`
`=(1.386 xx 10^(-3))/(2)=6.93 xx 10^(-4) mol min^(-1)`