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The reaction `Ararr B` follows first order kinetics. The time taken for `0.8 mol` of `A` to produce `0.6 mol` of `B` is `1 hr`. What is the time taken for the conversion of `9.0 mol` of `A` to Product `0.675 mol` of `B` ?A. `1 hr`B. `0.5 hr`C. `0.25 hr`D. `2 hr` |
Answer» Correct Answer - A For the reaction `2A + B rarr 3C + D` Rate constant of first-order reaction `k = (2.303)/(t) "log"_(10)((A)_(0))/((A)_(t))` or `k = (2.303)/(1) xx"log"_(10) (0.8)/(0.2)` …(i) (because `0.6` mole of `B` is formed) Suppose `t_(1)` and hour is required for changing the concentration of `A` from `0.9` mole to `0.675` mole of `B`. Remaining mole of `A 0.9 - 0.675 = 0.225` `:. k = (2.303)/(t) "log"_(10)(0.9)/(0.225)` Form Eqs (i) and (ii), we get `k = (2.303)/(t) "log"_(10)(0.8)/(0.2) = (2.303)/(t_(1)) "log"_(10)(0.9)/(0.225)` `2.303 log_(10) 4 = (2.303)/(t) log_(10) 4` `t_(1) = 1 hr` |
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