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InterviewSolution
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The activation energy of `H_(2)+I_(2) hArr 2HI(g)` in equilibrium for the forward reaction is `167 kJ mol^(-1)` whereas for the reverse reaction is `180 kJ mol^(-1)`. The presence of catalyst lowers the activation energy by `80 kJ mol^(-1)`. Assuming that the reactions are made at `27^(@)C` and the frequency factor for forwatd and backward reactions are `4xx10^(-4)` and `2xx10^(-3)` respectively, calculate `K_(c)`. |
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Answer» A catalyst lowers the activation energy for forward reaction as well as for backward reaction by equal amount. Thus, in presence of catalyst, Energy of activation for forward reaction `(E_(a1))=167-80=87 kJ "mol"^(-1)` Energy of activation for backward reaction `(E_(a2))=180-80=100 kJ "mol"^(-1)` `:.` For forward reaction, `K_(1)=A_(1)e^(-Ea//RT)` For backward reaction, `K_(2)=A_(2)e^(-Ea//RT)` where `A_(1)` and `A_(2)` are frequency factors and `E_(a1)` and `E_(a2)` are energies of activation after addition of catalyst. `:. K_(c )=K_(1)/K_(2)=A_(1)/A_(2) e^([(-E_(a1)//RT)+(-E_(a2)//RT)])` `=A_(1)/A_(2)e^(([E_(a2)-E_(a1)])/(RT))=(4xx10^(-4))/(2xx10^(-3)) e^(((100-87))/((8.314xx10^(-3)xx300)))` `:. K_(c )=2xx10^(-1) e^([13//8.314xx10^(-3)xx300])` `K_(c )=36.8` |
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