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Rate constant for the decomposition of ethylene oxide into `CH_(4)` and CO may be described be the equation. `logk(s)^(-1)=14.34-(1.25xx10^(4))/(T)` (a) What is the energy of activation of this reaction ? (b) What is the value of k at 670K ? |
Answer» (a) We know that, `log_(10)k=log_(10)A-(E)/(2.3030RT)" "....(i)` `log _(10)(s^(-1))=14.34-(1.25xx10^(4))/(T)" "....(ii)` Comparing eq. (i) and eq. (ii), we get `(E)/(2.303R)=1.25xx10^(4)` `E=1.25xx10^(4)xx2.303xx8.314xx10^(-3)` `E=239.339KJ//mol` (b) Substituting the value of T, i.e., at 670 K , in equation (ii), we get, `logk(s^(-1))=14.34-(1.25xx10^(4))/(670)=-4.3167` `k=4.82xx10^(-5)s^(-1)` |
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