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A first order reaction has `k=1.5 xx 10^(-6) s^(-1)` at `200^(@)`C. If the reaction is allowed to run for 10 hours, what percentage of initial concentration would have changed in products? |
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Answer» For the first order reaction, `t=2.303/k log a/(a-x)` `loga/(a-x) = (kxxt)/(2.303)=((1.5 xx 10^(-6)s^(-1)) xx (10 xx 60 xx 60s))/(2.303)=0.0235` `a/(a-x) ="Antilog" 0.0235=1.055` or `100=105.5 -1.055 x or x=(5.5)/(1.055) = 5.2=5.2%` |
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