Anonymous
Anonymous asked in Science & MathematicsChemistry · 7 years ago

The activation energy of a certain reaction is 49.6kJ/mol .?

The activation energy of a certain reaction is 49.6kJ/mol . At 23 ∘C , the rate constant is 0.0190s−1. At what temperature in degrees Celsius would this reaction go twice as fast?

Help me I've been stuck for awhile :(

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  • BII
    Lv 7
    7 years ago
    Favorite Answer

    Right from the start I know that rates typically double for every 10 °C rise in T so the answer should be ~33 °C

    Rate = Aexp-Ea1/RT

    ln(k2/k1) = (Ea/R)[1/T1 -1/T2]

    Ea in J mol^-1; R = 8.3145 J mol^-1;T in K; T1 =296 T2 to find;

    rate2/rate1 = 2.00

    ln(2.00) = 0.693 = [46600/8.3145](1/296 - 1/T2)

    (I tend to lose myself with these calc so I tread carefully!)

    0.693 ×8.3145/46600 = 1.237×10^-4 = 3.378×10^-3 -1/T2

    1/T2 = (3.378×10^-3) - (1.237×10^-4) = 3.254×10^-3

    T2 = 1/(3.254×10^-3) = 307 K or 34°C (Oh - you’re good drp :))

    http://www.chemguide.co.uk/physical/basicrates/arr...

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