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In a I order reaction `A rarr` products, the concentration of the reactant decrease to `6.25%` of its initial value in `80` minutes. What is (i) the rate constant and (ii) the rate of the reaction, `100` minutes after the start, if the initial concentration is `0.2 "mole"//"litre"` ?A. `2.17 xx 10^(-2) "min"^(-1), 3.47 xx 10^(-4) "mol.litre"^(-1) "min"^(-1)`B. `3.465 xx 10^(-2) "min"^(-1), 2.166 xx 10^(-4) "mol.litre"^(-1) "min"^(-1)`C. `3.465 xx 10^(-3) "min"^(-1), 2.17 xx 10^(-3) "mol.litre"^(-1) "min"^(-1)`D. ( d) `2.166 xx 10^(-3) "min"^(-1), 2.667 xx 10^(-4) "mol.litre"^(-1) "min"^(-1)` |
Answer» Correct Answer - B `K = (2.303)/(80)"log"((100)/(6.25)) = 3.465 xx 10^(-2)mm^(-1)` Then `3.465 xx 10^(-2) = (2.303)/(100)"log"((0.2)/(a_(t)))` `a_(t) = 0.00625` Rate `= K xx [a_(t)]` `= 0.00625 xx 3.465 xx 10^(-2)` `= 2.166 xx 10^(-4) sec^(-1)` |
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