The rate constant of a chemical reaction increased from 0.100 s?1 to 3.20s?1 upo

Question

The rate constant of a chemical reaction increased from 0.100 s?1 to 3.20s?1 upon raising the temperature from 25.0 ?C to 39.0?C .

Part A

Calculate the value of (1T2?1T1) where T1 is the initial temperature and T2 is the final temperature.

Part B

Calculate the value of ln(k1k2) where k1 and k2 correspond to the rate constants at the initial and the final temperatures as defined in part A.

Express your answer numerically.

Part C

What is the activation energy of the reaction?

Express your answer numerically in kilojoules per mole.

Explanation / Answer

k1 = 0.1/s.....T1 = 298K
k2 = 3.2/s.....T2 = 312K

k1/k2 = e^(-Ea/R)(1/T2 - 1/T1)
ln(k2/k1) = -Ea/R(1/T2 - 1/T1)
ln(3.2 / 0.1) = (-Ea/8.314J/mole-K)(1/312 - 1/298)
3.46 = -(Ea/8.314)(-1.505x10^-4)
-3.175x10^4 = -(Ea / 8.314)
1.91x10^5J or 191kJ

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