Search the web MasteringChemistry uNT 2 HOMEWORK Mozila Fir
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Solution
a)
we know,
ln (K2 / K1) = Ea / R * (T2 - T1) / T1 * T2
ln (0.227 / 1.23 * 10^-4) = Ea / 8.314 * (350 - 298) / 298 * 350
Ea = 125411.9 J / mole = 125.4 KJ / mole.
so activation energy is 125.4 KJ/ mole.
b)
At 16 oC = 273 + 16 = 289 K
ln (K2 / K1) = Ea / R * (T2 - T1) / T1 * T2
ln (1.23 * 10^-4 / K1) = 125411.9 / 8.314 * (298 - 289) / 289 * 298
K1 = 2.54 * 10^-5 s-1
