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A second order reaction in which both the reactants have same concentration , is 20% completed in 500 seconds. How much it will take for 60% completion ? |
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Answer» Solution :The second order equation when both the REACTANTS have same CONCENTRATION is `k=(1)/(t)[(1)/((a-x))-(1)/(a)]` If `a=100`, `x=20`, `t=500sec` `k=(1)/(500)xx(20)/(100xx(100-20))` When `a=100`, `x=60`, `t=?` `t=(1)/(k)[(1)/((100-60))-(1)/(100)]=(1)/(k)*(60)/(100xx40)` Substituting the value of `k` `t=(500xx100xx80)/(20)xx(60)/(100xx40)` `=3000` seconds. |
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