I need help with For Practice 14.8 . I\'m not sure how to calculate it without t

Question

I need help with For Practice 14.8.
I'm not sure how to calculate it without the frequency value, of even if that's needed. I did take the ln of one of the rates, 567 to get 6.3406. But again I'm wondering how to get the frequency value. Or am I suppose to use another formula.



Example 14.8 Using the Two-Point Form of the Arrhenius Equation Consider the reaction between nitrogendioxide and carbon monoxide: NO2 (g) CO(g) The rate constant at 701 K is measured as 2.57 M and that at 895 Kis measured as 567 M Find the activa tion energy for the reaction in kJ/mol. SORT You are given the rate constant of a reaction at two GIVEN: Ti 701 K, ki 2.57 M 1.s 1 different temperatures and asked to find the activation 895 K, k2 567 M energy FIND: E EQUATION In A2 E (1-1) STRATEGIZE Use the two-point form of the Arrhenius equation, which relates the activation energy to the given information and R (a constant). SOLVE Substitute the two rate constants and the two SOLUTION temperatures into the equation. 567 M E. 1 1 2.57 M 1.s R 1701 K 895 K EA 3.09 x 10 5.40 Solve the equation for Earthe activation energy, and convert to kJ/mol. E. 5.40 3.09 x 10 4 5.40 8.314 mol K 3.09 X 10 1.45 x 10 J/mol a 1.5 x 102 kl/mol CHECK The magnitude of the answer is reasonable. Activation energies for most reactions range from tens to hundreds of kilojoules per mole. FOR PRACTICE 14.8 Use the results from Example 14.8 and the given rate constant of the reaction at either of the two temperatures to predict the rate constant for this reaction at 525 K.

Explanation / Answer

Arhenius equation can be written as

lnK= lnko-Ea/RT, where K is rate constant ,Ko= Frequency factor, Ea= activation energy and T is temperature in K

Applying for two different temperatures

lnK1= lnKo - Ea/RT1 (1)

lnK2= lnKo- Ea/RT2 (2)

Eq.2- Eq.1 gives

ln(K2/K1)= (Ea/R)*(1/T1-1/T2)

T1= 710 K, K1= 2.57/M.s and T2= 895K, K2= 567/M.s

ln(567/2.57)= (Ea/8.314)*(1/710-1/895)

Ea= 154109 J/mole= 154 Kj/mole;

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