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The rate of a reaction gets doubled when the temperature changes from `7^(@)C` to `17^(@)C`. By what factor will it change for the temperature change from `17^(@)C` to `27^(@)C` ?A. `1.81`B. `1.71`C. `1.91`D. `1.76` |
Answer» Correct Answer - C `"log"(k_(2))/(k_(1)) = (E_(a))/(2.303R)[(1)/(T_(1)) - (1)/(T_(2))]` `log 2 = (E_(a))/(2.303R)[(T_(2) - T_(1))/(T_(1)T_(2))]` `0.3 = (E_(a))/(2.303R) xx (10)/(280 xx 290)` `Ea = 0.3 xx 2.303 xx 280 xx 29 xx R` …(1) Now again `"log"(k_(2))/(k_(1)) = (E_(a))/(2.303R)[(T_(2) - T_(1))/(T_(1)T_(2))]` `"log"(k_(2))/(k_(1)) = (0.3 xx 2.303 xx 280 xx 29 R)/(2.303 R)[(10)/(290 xx 300)] = 1.905` |
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