1.

For a decomposition reaction the values of rate constant k at two different temperatures are given below : `K_(1)=2.15xx10^(-8)L"mol"^(-1)s^(-1)"at " 650K` `K_(2)=2.39xx10^(-7)L"mol"^(-1)s^(-1)"at " 700K` Calclate the value of activation energy for this reaction. `(R=8.314JK^(-1)"mol"^(-1))`

Answer» Here `K_(1)=2.5xx10^(-8)L "mol"^(-1)S^(-1),K_(2)=2.39xx10^(-7)L" mol"^(-1)S^(-1)`
`T_(1)=650K,T_(2)=700 K and R=8.314JK^(-1)" mol"^(-1)`.
Using the formula,
`"log"(K_(2))/(K_(1))=(Ea)/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`"log" (2.39xx10^(-7))/(2.15xx10^(-8))=(Ea)/(2.303xx8.314)[(700-650)/(650xx700)]`
`"log" 1.11xx10=(Ea)/(19.147)xx(50)/(455000)`
`1.0457=(Ea)/(19.147)xx(1)/(9100)`
`:.Ea=1.0457xx19.147xx9100`
`=182200.36J`


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