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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 0.05 M, the rate of formation of O_(2) will be |
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Answer» `6.93xx10^(-2)" mol min"^(-1)` `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)` |
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