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The half life of a first order reaction x rarr products is 6.932 xx10^4 s at 500 K . What percentage of x would be decomposed on heating at 500 K for 100 min . (e^(0.06)=1.06) |
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Answer» Solution :Given `t_(1//2) = 0.6932xx10^4s` To solve : when t = 100 min , `([A_0]-[A])/([A_0])xx100= ? ` We known that a first order reaction , `t_(1//2)=(0.6932)/K` `k=(0.6932)/(6.932xx10^4)impliesk=10^(-5)s^(-1)` `k=(1/t)LN(([A_0])/([A]))` `10^(-5)s^(-1)xx100xx60s=ln([A_0])/([A])` `0.06=ln(([A_0])/([A]))` `([A_0])/([A])=E^(0.06)=1.06` `:. ([A_0]-[A])/([A_0])xx100%=(1-([A])/([A_0]))xx100%=(1-1/(1.06))xx100%=5.6%` |
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