%3Cp%3EBased%20upon%26nbsp%3Bthe%26nbsp%3Bdata%26nbsp%3Bbelow%2C%26nbsp%3Bdeterm
ID: 820268 • Letter: #
Question
%3Cp%3EBased%20upon%26nbsp%3Bthe%26nbsp%3Bdata%26nbsp%3Bbelow%2C%26nbsp%3Bdetermine%26nbsp%3Bthe%26nbsp%3Bactivation%26nbsp%3Benergy%26nbsp%3Bfor%26nbsp%3Bthe%26nbsp%3Breaction%26nbsp%3Band%26nbsp%3Bthe%26nbsp%3Bpreexponetial%26nbsp%3Bfactor%3A%3C%2Fp%3E%3Cp%3ET(Degrees%26nbsp%3BC)%26nbsp%3BRate%26nbsp%3Bconstant%26nbsp%3B(s%5E-1%5E)%3C%2Fp%3E%3Cp%3E0%26nbsp%3B9.16%26nbsp%3BE%26nbsp%3B-3%3C%2Fp%3E%3Cp%3E20%26nbsp%3B0.12919%3C%2Fp%3E%3Cp%3E40%26nbsp%3B1.2993%3C%2Fp%3E%3Cp%3E50%26nbsp%3B3.7017%3C%2Fp%3E%3Cp%3E70%26nbsp%3B25.017%3C%2Fp%3E%3Cp%3E%3Cbr%3E%3C%2Fp%3EExplanation / Answer
log(K2/K1) = Ea/2.303R[1/T1 -1/T2]
log(1.2993/0.12919) = Ea /2.303*8.314 [ 1/293 - 1/313 ]
Ea = 88016.32 J or 88.02 KJ
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