Activation Energy and Arrhenius Equation
Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 49 to 59°C?
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49 deg = 49 + 273 = 322 K
59 deg = 59 + 273 = 332 K
We have the Arrhenius equation: k = A exp[-Ea/RT]
Take log
ln k = ln A - Ea/(RT)
Let k1 and k2 ...
Solution Summary
The solution includes a calculation of activation energy using the Arrhenius equation, given the change in temperature of a solution.
Arrhenius equation, entropy, enthalpy and Gibbs free energy
For the reaction:
H2 + C2H4 -> C2H6
In gaseous phase, the pre-exponential factor is 1.24*10^6 [M-1s-1] and the activation energy is 180 kJ/mol. find the entropy, the enthalpy and the gibbs free energy.
see attachment
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