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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 beA. `6.93 xx 10^(-4) mol "min"^(-1)`B. `2.66 L "min"^(-1)` at STPC. `1.34 xx 10^(-2) mol "min"^(-1)`D. `6.93 xx 10^(-2) mol "min"^(-1)` |
Answer» Correct Answer - a 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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