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A first order reaction is 50% completed in 20 minutes at `27^(@)`C. The energy of activation of the reaction is-A. 43.58 KJB. 55.14 KJC. 11.97 KJD. 6.65 KJ |
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Answer» Correct Answer - B `"Log"((K_(2))/(K_(1)))=(Ea)/(2.303R)((1)/(T_(1))-(1)/(T_(2)))` `Log4=(Ea)/(2.303xx8)((1)/(300)-(1)/(320))` `{K_(1)=(1n2)/(20),K_(2)=(n2)/(5)}` |
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