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The activation energy of a reaction is 9 kcal `"mole "^(-1) ` . The increase in the rate cnstant when its temperature is raised from 295 to 300 approximatelyA. `1.289` timesB. `12.89` timesC. `0.1289` timesD. 0.25 |
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Answer» Correct Answer - A `log ""(K_(2))/(K_(1)) =(E_(a). DT )/(2.303RT_(2)T_(1))` `=(9000xx5)/(2.303xx2xx300xx295)=0.1104` `log ""(k_(2))/(k_(1))=0.1104 therefore (K_(2))/(K_(1))=1.289` ` therefore K_(2)=K_(1)xx1.289` |
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