Use the integrated form of the Arrhenius equation as ln k 1 /k 2 = -E a /R*(1/T

Question

Use the integrated form of the Arrhenius equation as

ln k1/k2 = -Ea/R*(1/T1 – 1/T2)

where k1 = 0.005 rate units is the rate constant at temperature T1 = 330 K and k2 = 0.001 rate units is the value of the rate constant at temperature T2 = 310 K. Plug in values and obtain

ln (0.005/0.001) = -Ea/(8.314 J/mol.K)*[(1/330 K) – (1/310 K)]

====> ln 5 = -Ea/(8.314 J/mol.K)*(0.003030 – 0.003226) K-1

====> 1.6094 = -Ea/(8.314 J/mol)*(-0.000196)

====> Ea = 1.6094*(8.314 J/mol)/(0.000196) = 68268.120 J/mol = (68268.120 J/mol)*(1 kJ/1000 J) = 68.268120 kJ/mol 68.2681 kJ/mol (ans).

The rate constants are low and the reaction is therefore, most likely chemical. However, a comparison of rate constants for two or more steps of a reaction is required to answer the question completely.

Explanation / Answer

2. (20) Given the following values of a specific rate constant determined from two kinetic experiments: kTCK) 0.005 10 330 0.001 And the Arrhenius Equation: R987mea 14 mol K mol K Determine the activation energy (Ea) of the reaction. Can you predict if the reaction is chemical or diffusion limited?

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.