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The decomposition of A into products has a value of k as 4.5 xx 10^(3) s^(-1) at 10^(@)C and energy of activation 60 kJ mol^(-1). At what temperature would k be 1.5 xx 10^(-4) s^(-1) ? |
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Answer» Solution :Given, `T_(1) = 10^(@)C + 273 = 283 K` `k_(T_(1)) = 4.5 xx 10^(3) sec^(-1)` `E_(a) = 60 KJ mol^(-1) =6 xx 10^(3) J mol^(-1)` `T_(2) = ?` `k_(T_(2)) = 1.5 xx 10^(4) sec^(-1)` We know that `log.(k_(T_(1))/(k_(T_(1)))) = (E_(a))/(2.303 R)[(T_(2)-T_(1))/(T_(1) xx T_(2))]` On putting values `log.(1.5 xx 10^(4))/(4.5 xx 10^(3)) = (60 xx 10^(3) J mol^(-1))/(2.303 xx 8.314)((T_(2)- 283)/(283 T_(2)))` `0.523 = 3133.63((T_(2)-283)/(283 T_(2)))` `(T_(2) - 283)/(283 T_(2)) = 1.67 xx 10^(-4)` `T_(2) - 283 = 0.0472 T_(2)` `0.953 T_(2) = 283` `T_(2) = 297 K` `T_(2) = 24^(@) C` |
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