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In a reaction, `n_(1)A+n_(2)B rarr m_(1)C+m_(2)D, 5mol litre^(-1)` of `A` are followed to react with `3 mol litre^(-1)` of `B`. After `5` second, the concentration of `A` was found to be `B`. After `5` second, the concentration of `A` was found to be `4M`. Calculate rate of reaction in terms of `A` and `D`. |
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Answer» Change in `[A]` in `5 sec =5M-4M =1M` `:.` Rate of reaction in terms of `A=-(d[A])/(dt)` `=1/5 =0.2 M sec^(-1)` Also, `=1/n_(1)(d[A])/(dt)=1/m_(2)(d[D])/(dt)` `:.` Rate of reaction in terms of `D` `=(d[D])/(dt)=m_(2)/n_(1)xx0.2 M sec^(-1)` |
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