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How much faster would a reaction proceed at `25^(@)C` than at `0^(@)C` if the activation energy is `65 kJ` ?A. `2` timesB. `5` timesC. `11` timesD. `16` times |
Answer» Correct Answer - C `"log"((K_(2))/(K_(1))) = (E_(a))/(2.303R)((T_(2) - T_(1)))/(T_(1)T_(2))` `= (65 xx 10^(3)xx(298-273))/(2.303 xx 8.3 xx 298 xx 273)` By calculation we fins `(K_(2))/(K_(1)) = 11` |
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