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. `6.93 xx 10^(-2)` mol`min^(-1)`
B. `6.93 xx 10^(-4)mol L^(-1)min^(-1)`
C. `2.66 L min^(-1)` at STP
D. `1.34 xx 10^(-2)mol min^(-1)`

Answer

B

Explanation

b) `H_(2)O_(2) to H_(2)O + 1/2O_(2)`
In 50 minutes , concentration of `H_(2)O_(2)` decreases from 0.5 M to 0.125 M i.e. gets reduced to `1/4`

`2t 1//2=50"min",t_(1//2)=25"min"`
`k=(0.693)/(t_(1//2))=0.693/25min^(-1)`
Rate of decomposition of `H_(2)O_(2)=k[H_(2)O]`
`=(0.693)/(25) min^(-1)(0.05 mol L^(-1))`
`O_(2) = 1.386 xx 10^(-3) mol L^(-1)min^(-1)`
Rate of formation of =`1/2` x Rate of decomposition of `K_(2)O_(2)`
`=1/2 xx 1.386 xx 10^(-3)`
`=6.93 xx 10^(-4)"mol" L^(-1)min^(-1)`