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The rate constant for a first order reaction is 60 s^(-1). How much time will it take to reduce 1 g of the reactant to 0.0625 g? |
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Answer» Solution :The EQUATION for a first order reaction is : `t=(2.303)/(k)"LOG"([R]_(0))/([R])` `[R]_(0)=1 g, [R]=0.0625 g, k=60s^(-1)` Substituting the values in the first order equation, we have `t=(2.303)/(60) "log"(1)/(0.0625)=(2.303)/(60)log 16=(2.303)/(60)xx1.204=0.0462s` |
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