Ea  of an uncatalyzed reaction is 50kj/mole .addition of a catalyst increases the rate 10^5 times at 25 celcius.calculate ea of catalyzed reaction

Asked by raman.mathur100 | 10th Oct, 2015, 11:59: PM

Expert Answer:

begin mathsize 14px style Consider space uncatalysed space as space straight k space and space catalysed space rate space constant space will space be space straight k to the power of apostrophe. So comma space straight k to the power of apostrophe space equals space straight k space straight x space 10 to the power of 5 Let space straight E subscript straight a space end subscript be space the space activation space energy space for space uncatalysed space reaction space and space straight E subscript straight a superscript apostrophe space be the space activation space energy space for space catalysed space reaction. log space straight k space equals space log space straight A space minus space begin inline style fraction numerator straight E subscript straight a space end subscript over denominator 2.303 space RT end fraction end style  log space straight k apostrophe space equals space log space straight A space minus space begin inline style fraction numerator straight E subscript straight a superscript apostrophe over denominator 2.303 space RT end fraction end style But space space straight k to the power of apostrophe space equals space straight k space straight x space 10 to the power of 5 So comma log space straight k space straight x space 10 to the power of 5 space equals space log space straight A space minus space begin inline style fraction numerator straight E subscript straight a superscript apostrophe over denominator 2.303 space RT end fraction end style  log space straight k space plus space 5 space log space 10 space equals space log space straight A space minus space begin inline style fraction numerator straight E subscript straight a superscript apostrophe over denominator 2.303 space RT end fraction end style log space straight A space minus space begin inline style fraction numerator straight E subscript straight a space end subscript over denominator 2.303 space RT end fraction end style space plus space 5 space equals log space straight A space minus space begin inline style fraction numerator straight E subscript straight a superscript apostrophe over denominator 2.303 space RT end fraction end style  minus space begin inline style fraction numerator straight E subscript straight a space end subscript over denominator 2.303 space RT end fraction end style space space plus space fraction numerator straight E subscript straight a superscript apostrophe over denominator 2.303 space RT end fraction space equals space minus 5  space fraction numerator straight E subscript straight a superscript apostrophe space minus space 50 over denominator 2.303 space RT end fraction space space equals space minus 5 space straight x space 2.303 space straight x space 8.314 space straight x space 10 to the power of negative 3 end exponent space straight x space 298  straight E subscript straight a superscript apostrophe space equals space 21.47 space KJ space divided by space mole space       end style

Answered by Arvind Diwale | 12th Oct, 2015, 06:45: PM