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The rate constant 'k'. For a reaction varies with temperature 'T' according to the question. log k = log A-(E_(a))/(2.303R)(1/T) Where E_(a) is the activation energy. When a graph is plotted for log k vs 1//T, a straight line with a slope of -4250 K is obtained. Calcualte E_(a) for this reaction. |
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Answer» SOLUTION :SLOPE =`E_a/(2.303R)`=-4250 K So, `E_a=-2.303xxRxx"Slope"` `=-2.303xx8.314 J K^(-1) "mol"^(-1) xx4250` `=81375.3 "J mol"^(-1)` `=81.375 "KJ mol"^(-1)` |
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