The activation energy for the following reaction: Sn2+ + 2 Co3+→ Sn4+ + 2 Co2+ is 60.0 kJ/mol.
By what factor (how many times) will the rate constant increase when the temperature is raised from 10°C to 28°C?
Hint: Consider which temperature: lower or higher corresponds bigger rate constant.
Example of answer: = [type your answer]. (k2 > k1)
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The activation energy for the following reaction: Sn2+ + 2 Co3+→ Sn4+ + 2 Co2+ is...
Calculate the activation energy, in kJ/mol, for the redox reaction Sn2+ + 2Co3+ ----> Sn4+ + 2Co2+ T (˚C) k (M-1s-1) -1 3.12 x 103 62 30.0 x 103 A. -0.269 B. -21.9 C. 59.2 D. 0.269 E. -27.2 F. 21.9 G. 27.2 H. answer not listed
Learning Goal: To use the Arrhenius equation to calculate the activation energy. As temperature rises, the average kinetic energy of molecules increases. In a chemical reaction, this means that a higher percentage of the molecules possess the required activation energy, and the reaction goes faster. This relationship is shown by the Arrhenius equation k=Ae−Ea/RT where k is the rate constant, A is the frequency factor, Ea is the activation energy, R = 8.3145 J/(K⋅mol) is the gas constant, and T...
Using the Arrhenius equation to calculate the activation energy. The rate constant of a chemical reaction increased from 0.100s-1 to 2.90s-1 upon raising the temperature from 25 to 45 C (1/t2 -1/t1)= -2.11x10^-4 K-1 Calculate the value of In (k1/k2) where k1 and k2 corresponds to the rate constant at the initial and the final temperature as found above. In(k1/k2)=?? Also, what is the activation energy of the reaction? Expressed in kilojoules per mile Ea=??
Part A: The activation energy of a certain reaction is 45.9 kJ/mol . At 27 ∘C , the rate constant is 0.0130s−1 . At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units. Part B: Given that the initial rate constant is 0.0130s−1 at an initial temperature of 27 ∘C, what would the rate constant be at a temperature of 130. ∘C for the same reaction described in Part A? Part C:...
To use the Arrhenius equation to calculate the activation energy. As temperature rises, the average kinetic energy of molecules increases. In a chemical reaction, this means that a higher percentage of the molecules possess the required activation energy, and the reaction goes faster. This relationship is shown by the Arrhenius equation k=Ae−Ea/RT where k is the rate constant, A is the frequency factor, Ea is the activation energy, R = 8.3145 J/(K⋅mol) is the gas constant, and T is the...
To use the Arrhenius equation to calculate the activation energy. As temperature rises, the average kinetic energy of molecules increases. In a chemical reaction, this means that a higher percentage of the molecules possess the required activation energy, and the reaction goes faster. This relationship is shown by the Arrhenius equation k=Ae−Ea/RT where k is the rate constant, A is the frequency factor, Ea is the activation energy, R = 8.3145 J/(K⋅mol) is the gas constant, and T is the...
Part A The activation energy of a certain reaction is 30.9 kJ/mol . At 29 ∘C , the rate constant is 0.0170s−1 . At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units. View Available Hint(s) T2 = Part B Given that the initial rate constant is 0.0170s−1 at an initial temperature of 29 ∘C , what would the rate constant be at a temperature of 100. ∘C for the...
For a parallel reaction A goes to B with rate constant k1 and A goes to C with rate constant k2, you determine that the activation energies are 38.3 kJ/mol for k1 and 169.3 kJ/mol for k2. If the rate constants are equal at a temperature of 316 K, at what temperature (in K) will k1/k2 = 2?
Use the Arrhenius equation to calculate the activation energy. The rate constant of a chemical reaction increased from 0.100 s−1 to 2.70 s−1 upon raising the temperature from 25.0 ∘C to 43.0 ∘C . a) Calculate the value of (1/T2−1/T1) where T1 is the initial temperature and T2 is the final temperature. (in units of k-1) b) Calculate the value of ln(k1/k2) where k1 and k2 correspond to the rate constants at the initial and the final temperatures as defined...
A certain reaction has an activation energy of 60.0 kJ/mol and a frequency factor of A1 = 2.10×1012 M−1s−1 . What is the rate constant, k, of this reaction at 29.0 ∘C ? Express your answer with the appropriate units. Indicate the multiplication of units explicitly either with a multiplication dot (asterisk) or a dash.