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InterviewSolution
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The activation energy of `H_(2)+I_(2) hArr 2HI` in equilibrium for the forward reaction is `167 kJ mol^(-1)` whereas for the reverse reaction is `180 kJ mol^(-1)`. The presence if catalyst lowers the activation energy by `80 kJ mol^(-1)`. Assuming that the reaction are made at `27^(@)C` and the frequency factorr for the forward and backward reactions are `6xx10^(-4)` and `3xx10^(-3)`, respectively, calculate `K_(c )`. |
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Answer» The lowering of activation energy by a catalyst occurs for forward reaction as well as for backward reaction. Thus, in presence of catalyst, Energy of activation for forward reaction `(DeltaH_(1))` `=167-80=37 kJ "mole"^(-1)` Energy of activation for backward reaction `(DeltaH_(2))` `=180-80=100 kJ "mole"^(-1)` For forward reaction, `K_(1)=A_(1) e^(-DeltaH_(1)//RT)` For backward reaction `K_(2)=A_(2) e^(-DeltaH_(2)//RT)` where `A_(1)` and `A_(2)` are frequency factors and `DeltaH_(1)` and `DeltaH_(2)` are energies of activation. `:. K_(c )=K_(1)/K_(2)=A_(1)/A_(2). e^([(-DeltaH_(1)//RT)+(DeltaH_(2)//RT)])` `=(6xx10^(-4))/(3xx10^(-3))xxe^([(-87+100)//(8.314xx10^(-3)xx300)])` `K_(c )=2xx10^(-1) e^(+13//(8.314xx300xx10^(3)))=36.8` |
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