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

Answer

B

Explanation

Decompostion of `H_(2)0_(2)` takes place as :
`H_(2)O_(2) to H_(2)O + (1)/(2)IO_(2)`
Decrease of concentration of `H_(2)O_(2)` from 0.5 to 0.125 in 50 minutes means two half lives `(0*5 to 0*25 to 0*125)`
i.e. `2xxt_(1//2) = 50 min or t_(1//2) = 25 min`
For a 1st order reaction,
`k = (0*693)/(t_(1//2)) = (0*693)/(25) min^(-1)`
Rate of reaction = Rate of decomposition of
`H_(2)O_(2),` i.e., `(d[H_(2)O_(2)])/(dt) = k [H_(2)O_(2)]`
`= (0*693)/(25) xx 0*05 = 1*386 xx 10^(-3) mol L^(-1) min^(-1)`
Rate of formation of `O_(2) = (d[O_(2)])/(dt) = -(1)/(2)(d[H_(2)O_(2)])/(dt)`
` = (1)/(2) xx 1*386 xx 10^(-3)`
`= 6*93 xx 10 ^(-4) mol L^(-1) min^(-1)`