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In general it is observed that the rate ...

In general it is observed that the rate of a chemical reaction becomes double for every `10^@` rise in temperature. If this generalisation holds for a reaction in the temperature range 2908 K to 398 K, what would be the value of activation energy for the reaction.
(R=8.314 J `K^(-1)mol^(-1)`)

Text Solution

Verified by Experts

`"log" (k_2)/(k_1) = (E_a)/(2.303) [1/(T) - 1/(T_2)] " " [A.E.]`
`(k_2)/(k_1) =2 `(given in question)
`log 2 = (E_a)/(2.303 xx 8.314)[1/298 - 1/308]`
`E_a = (0.3010 xx 2.303 xx 8.314 xx 298 xx 308)/(10)`
`E_a = 52898 J mol `
`E_a = 52.898 KJ mol^(-1)`
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