InterviewSolution
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InterviewSolution
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The ionization constant of `overset(o+)(NH_(4))` ion in water is `5.6 xx 10^(-10)` at `25^(@)C`. The rate constant the reaction of `overset(o+)NH_(4)` and `overset(ɵ)(OH)` ion to form `NH_(3)` and `H_(2)O` at `25^(@)C` is `3.4 xx 10^(10)L mol^(-1) s^(-1)`. Calculate the rate constant for proton transfer form water to `NH_(3)`. |
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Answer» Correct Answer - Rate constant `= 6.07 xx 10^(5) s^(-1)` `NH_(3) + H_(2)O underset(k_(b))overset(k_(f))hArr NH_(4)^(o+)+overset(Θ )(OH) (k_(b) = 3.4 xx 10^(10))` ...(i) `NH_(4)^(o+) + H_(2)O hArr NH_(4)OH +H^(o+) (k_(a) = 5.6 xx 10^(-10))` `k_(base)(NH_(3))=(k_(f))/(k_(b))=(k_(w))/(k_(acid)(NH_(4)^(o+)))` `( :. k_(acid)xxk_(base)=k_w)` `(k_(f))/(3.4xx10^(10))=(10^(-14))/(5.6xx10^(-10))impliesk_(f)= 6.07xx10^(5)` (Alternative methof) We are given that `NH_(4)^(o+) +H_(2)O hArr NH_(3) + H_(3)O^(o+) (k_(a) = 5.6 xx 10^(-10))` `NH_(4)^(o+) + overset(Θ)(OH) underset(k_(b))overset(k_(f))hArr NH_(3) + H_(2)O (k_(f) = 3.4 xx 10^(10) L mol^(-1) s^(-1))` The second equation can be generalized as followes: `{:(NH_(4)^(o+) + H_(2)O hArr NH_(3) + H_(3)O^(o+) (k_(a))),(" "2H_(2)O hArr H_(3)O^(o+) + overset(Θ )(OH) (k_(w))),(ul(bar("Subtract" : NH_(4)^(o+) + overset(Θ )(OH) hArr NH_(3)+H_(2)O))):}` ...(ii) `k_(eq)=(k_(a))/(k_(w))` and also `k_(eq)=(k_(f))/(k_(b))` `:. (k_(f))/(k_(b))=(k_(a))/(k_(w))` `k_(b)=k_(f)[["Note here that equation(ii)is reversed",],["of Eq.(i) So" k_(b)"is calculated",]]` `=(3.4xx10^(10)Lmol^(-1)s^(-1))((1.0xx10^(-14)mol^(2)L^(-2))/(5.6xx10^(-10)molL^(-1)))` `= 6.07xx10^(5)s^(-1)`Correct Answer - Rate constant `= 6.07 xx 10^(5) s^(-1)` `NH_(3) + H_(2)O underset(k_(b))overset(k_(f))hArr NH_(4)^(o+)+overset(Θ )(OH) (k_(b) = 3.4 xx 10^(10))` ...(i) `NH_(4)^(o+) + H_(2)O hArr NH_(4)OH +H^(o+) (k_(a) = 5.6 xx 10^(-10))` `k_(base)(NH_(3))=(k_(f))/(k_(b))=(k_(w))/(k_(acid)(NH_(4)^(o+)))` `( :. k_(acid)xxk_(base)=k_w)` `(k_(f))/(3.4xx10^(10))=(10^(-14))/(5.6xx10^(-10))impliesk_(f)= 6.07xx10^(5)` (Alternative methof) We are given that `NH_(4)^(o+) +H_(2)O hArr NH_(3) + H_(3)O^(o+) (k_(a) = 5.6 xx 10^(-10))` `NH_(4)^(o+) + overset(Θ)(OH) underset(k_(b))overset(k_(f))hArr NH_(3) + H_(2)O (k_(f) = 3.4 xx 10^(10) L mol^(-1) s^(-1))` The second equation can be generalized as followes: `{:(NH_(4)^(o+) + H_(2)O hArr NH_(3) + H_(3)O^(o+) (k_(a))),(" "2H_(2)O hArr H_(3)O^(o+) + overset(Θ )(OH) (k_(w))),(ul(bar("Subtract" : NH_(4)^(o+) + overset(Θ )(OH) hArr NH_(3)+H_(2)O))):}` ...(ii) `k_(eq)=(k_(a))/(k_(w))` and also `k_(eq)=(k_(f))/(k_(b))` `:. (k_(f))/(k_(b))=(k_(a))/(k_(w))` `k_(b)=k_(f)[["Note here that equation(ii)is reversed",],["of Eq.(i) So" k_(b)"is calculated",]]` `=(3.4xx10^(10)Lmol^(-1)s^(-1))((1.0xx10^(-14)mol^(2)L^(-2))/(5.6xx10^(-10)molL^(-1)))` `= 6.07xx10^(5)s^(-1)` |
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