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The energy of activation for a first ...

The energy of activation for a first order reaction is 104 kJ `mol^(-1) ` The rate constant at `25^(@) C` is `3.7 xx 10^(-5) s^(-1)` . What is the rate constant at `30^(@) C`

Text Solution

Verified by Experts

The correct Answer is:
Rate constant at `30^(@) C = 7.4 xx 10^(-4) s^(-1)`

Given : Energy of activation `=E_(a) =104 kJ mol^(-1) = 104 xx 10^(3) j mol^(-1)`
Initial rate constant = `k_(1) = 3.7 xx 10^(-5) s^(-1)`
Initial temperature = `T_(1)= 273 + 25 = 298 K`
Final temperature =`T_(2) = 273 + 30 = 303 K`
Final rate constant = `k_(2) = ?`
`log_(10) .(k_(2))/(k_(1)) = (E_(a) (T_(2)- T_(1)))/(2.303 R xx T_(1) xx T_(2))`
`= (104xx 10^(3) xx (303 - 298))/(2.303 xx 8.314 xx 298 xx 303) = (104 xx 10^(5) xx 5)/(2.303 xx 8.314 xx298 xx 303) =0.3`
`:. (k_(2))/(k_(1)) = AL 0.3 = 1.995 ~= 2`
`:. k_(2) = 2 xx k_(1) = 2xx 3.7 xx 10^(-5) = 7.4 xx 10^(-4) s^(-1)`
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