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[6]BONUS: kl -5,5x10-3 3-1 at 315 K, find k2 at 415 Kif Ea - 45.5 kJ/mol.
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Initial conditions k = 5.5 X 10 S-1 Ti = 315k Final conditions K₂ = ? I2 = 4isk Eg= 45.5kJ/mol Ea = 45.5 8 10 ² J. mat Univer

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[6]BONUS: kl -5,5x10-3 3-1 at 315 K, find k2 at 415 Kif Ea - 45.5 kJ/mol.
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