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It is generalized that a `10^(@)C` increases in temperature casues the rate of reaction to double. Applied to a reaction at `295 K`, what is the value of `E_(a)`?A. `120 "kcal mol"^(-1)`B. `1200 "kcal mol"^(-1)`C. `1.2 "kcal mol"^(-1)`D. `12 "kcal mol"^(-1)` |
Answer» Correct Answer - D Only reaction whose `E_(a)` falls in the range of `50-55 kJ mol^(-1)` or `12-13 kcal mol^(-1)` are found to double their rate for `10^(@)C` rise in temperature i.e., form `298` to `308 K`. Alternatively Use Arrheinius equation `T_(1) = 295 K, T_(2) = 305 K`. `log.(k_(2))/(k_(1)) = (E_(a))/(2.3 xx 2 xx 10^(-3))((10)/(295 xx 305))` `E_(a) = (log(2) xx 2.3 xx 2 xx 10^(-3) xx 295 xx 305)/(10)` `E_(a) = (0.3 xx 4.6 xx 295 xx 305)/(10^(4))` `= 12.4 kcal mol^(-1)` `~~ 12 kcal mol^(-1)`Correct Answer - D Only reaction whose `E_(a)` falls in the range of `50-55 kJ mol^(-1)` or `12-13 kcal mol^(-1)` are found to double their rate for `10^(@)C` rise in temperature i.e., form `298` to `308 K`. Alternatively Use Arrheinius equation `T_(1) = 295 K, T_(2) = 305 K`. `log.(k_(2))/(k_(1)) = (E_(a))/(2.3 xx 2 xx 10^(-3))((10)/(295 xx 305))` `E_(a) = (log(2) xx 2.3 xx 2 xx 10^(-3) xx 295 xx 305)/(10)` `E_(a) = (0.3 xx 4.6 xx 295 xx 305)/(10^(4))` `= 12.4 kcal mol^(-1)` `~~ 12 kcal mol^(-1)` |
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