Question

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A researcher raises the temperature from 44.2 to 61.6 °C and finds that the rate of the reaction doubles. What was the activation energy in /mol) for this reaction? (R-8.3145 /mold mol

Answer 1

Know the tonmuha We Ea K2 ;T2T 2.303 R Whene, tempenahere T tampenalure Rote constom a T K2 Ki Rate constomt o Ea- Activation eneng 11 Univensok Gon conshomt R =

Griven K1 K CSupposa) Then, K=2K Accerding to question) 1 T,61.6C= (G1.6+273) K = 334.6 K 44.2 C=(44.2 273)K 317.2 K = 8.3145 J.mol. K. Civen) R Ea 2 1

Accondingy Ea K2 X 2.303 xR Ea x [수-1 K2 X 2,303 xR or, dog or, dog 1 2K 2.303 x 8.3145 3.mol-. K-' Ea x 334.6 K 317.2 K Or, dog 2 x 2.303 x 8.31453. mol. K Eax 317.2 334.6 Eax (s.64 x10-4) K- 76 J.mo-. K-! 5.76 J. Mof 64 x10 or, Ea 35121.95 J. mof Ea OY, 35121.95 KJ. mol Ea = or, -35.12195 KJ.mofI 1000 35.12 KJ. mof The vohue o Ea in 35.12 KJ/mot. Ang .