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InterviewSolution
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At `380^(@)C` , the half-life periof for the first order decompoistion of `H_(2)O_(2)` is `360 min`. The energy of activation of the reaction is `200 kJ mol^(-1)`. Calculate the time required for `75%` decompoistion at `450^(@)C`. |
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Answer» Correct Answer - `20.358 min` The first order reaction Half life `=(0.693)/(k)` or `360=(0.693)/(k)` `k_(380^(@)C)=(0.693)/(360)` `k=Ae^(-E_(a)//RT)` or `log k = log A-(E_(a))/(2.303RT)` `:. log.(k_(2))/(k_(1))= (E_(a))/(2.303TR)[(1)/(T_(1))-(1)/(T_(2))]` or `log.(k_(450^(@)C))/((0.693)/(360))=(200xx1000)/(2.303xx8.314)[(1)/(653)-(1)/(723)]` or `k_(450^(@)C)=6.18xx10^(2-)min^(-1)` For `75%` decompoistion at `723 K` `k_(450^(@)C)= (2.303)/(t) log.(a)/(a-x)` or `6.81xx10^(-2)=(2.303)/(t)log.(100)/(25)` or `t= 20.358min`Correct Answer - `20.358 min` The first order reaction Half life `=(0.693)/(k)` or `360=(0.693)/(k)` `k_(380^(@)C)=(0.693)/(360)` `k=Ae^(-E_(a)//RT)` or `log k = log A-(E_(a))/(2.303RT)` `:. log.(k_(2))/(k_(1))= (E_(a))/(2.303TR)[(1)/(T_(1))-(1)/(T_(2))]` or `log.(k_(450^(@)C))/((0.693)/(360))=(200xx1000)/(2.303xx8.314)[(1)/(653)-(1)/(723)]` or `k_(450^(@)C)=6.18xx10^(2-)min^(-1)` For `75%` decompoistion at `723 K` `k_(450^(@)C)= (2.303)/(t) log.(a)/(a-x)` or `6.81xx10^(-2)=(2.303)/(t)log.(100)/(25)` or `t= 20.358min` |
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