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The rate of formation of `C_(6)H_(6)+3H_(2) underset(k_(b))overset(k_(f))hArr C_(6)H_(12)` for the forward reaction is first order with respect to `C_(6)H_(12)` and `H_(12)` each. Which one of the options is/are correct?A. `k_(eq) = (k_(f))/(k_(b))`B. `k_(eq) = ([C_(6)H_(12)])/([C_(6)H_(6)][H_(2)]^(3))`C. `r_(f) =k_(f)[C_(6)H_(6)][H_(2)]`D. `r_(b) = k_(b)[C_(6)H_(12)][H_(2)]^(-2)` |
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Answer» Correct Answer - A::B::C::D `k_(eq) = (k_(f))/(k_(b)) = ([C_(6)H_(12)])/([C_(6)H_(6)][H_(2)]^(3)) = ([C_(6)H_(12)])/([C_(6)H_(6)][H_(2)][H_(2)]^(2))` `r_(f) = k_(f) xx[C_(6)H_(6)][H_(2)]` `r_(b) = k_(b) xx "Unknown"` At equilibrium, `r_(f) = r_(0)` `:. k_(f)[C_(6)H_(6)][H_(2)] = k_(b) xx " Unknown"` `r_(b) = k_(b) xx (k_(f))/(k_(b)) xx [C_(6)H_(6)][H_(2)]` `= k_(f)[C_(6)H_(6)][H_(2)]` `= k_(b)[C_(6)H_(12)][H_(2)]^(-2)` |
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