1.

Decomposition of H_(2)O_(2) follows a first order reaction . In fifty minutes 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 from 0.05 M , the rate of formation of O_(2) will be

Answer»

`6.93 xx 10^(-4) mol "min"^(-1)`
`2.66 L "min"^(-1)` at STP
`1.34 xx 10^(-2) mol "min"^(-1)`
`6.93 xx 10^(-2) mol "min"^(-1)`

Solution :In 50 minutes , concentration of `H_(2)O_(2)` becomes `(1)/(4)` of initial
`implies 2 xx t_(1//2) = 50 ` minutes `"" t_(1//2) = 25` minutes
`implies K = (0.693)/(25) ` PER minute
`r_(H_(2) O_(2)) = (0.693)/(25) xx 0.05 = 1.386 xx 10^(-3)`
`2H_(2)O_(2) to 2H_(2)O + O_(2) "" r_(O_(2)) = (1)/(2) xx r_(H_(2)O_(2))`
`r_(O_(2)) = 0.693 xx 10^(-3) "" r_(O_(2)) = 6.93 xx 10^(-4)` mol/minute `xx` LITRE


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