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For reaction `AtoB`, the rate constant `k_(1)=A_(1)e^(-Ea)//RT` and for the reaction `PtoQ`, the rate constant `k_(2)=A_(2)e^(-Ea_(2))//RT`. If `A_(1)=10^(8),andEa_(1)=600"cal/mol",Ea_(2)=1200"cal/mol"`, then the temperature at which `k_(1)=k_(2)` is (R=2cal/K-mol)A. 600 KB. `300xx4.606` KC. `(300)/(4.606)K`D. `(4.606)/(600)K` |
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Answer» Correct Answer - C `A_(1)e-Ea_(1)//RT=A_(2)e-Ea_(2)//RT` `(A_(2))/(A_(1))=e(Ea_(2)-Ea_(1))//RT` `10^(2)="Exp"{(600)/(RT)}" R=2cal/K-mol"` `2" ln "10=(600)/(2T)` `T={(300)/(2xx2.303)}K" "]` |
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