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ID: 1039695 • Letter: I
Question
I need help with this question please help me with it and be clear to understand and clear to see and read it It is important for my chemistry homework please do it and show correct and clear steps so I can do it also e clear sometimes it is not clear please be clear
10. At 273 K the rate constant for the decomposition of NO2 is 7.78 x 10. 2 NO2 (g) N2 (g)+202 (g) kJ/mole. At 298 K the rate constant is 3.46 x 10-7. The energy of activation for this reaction is Show work: Equation: Numerical Substitution Final answer with correct units and significant digits:Explanation / Answer
Equation is:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
Given:
T1 = 273 K
T2 = 298 K
K1 = 7.78*10^-7
K2 = 3.46*10^-7
use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(3.46*10^-7/7.78*10^-7) = ( Ea/8.314)*(1/273.0 - 1/298.0)
-0.8103 = (Ea/8.314)*(3.073*10^-4)
Ea = -21922 J/mol
Ea = -21.922 KJ/mol
Answer: -21.9 KJ/mol