Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calculate the value of the rate constant at 25 °C. Alternatively, you can simply extrapolate the straight line plot of ln(k) vs. 1/T in your notebook to 1/298 , read off the value of ln(k), and determine the value of k. Please put your answer in scientific notation. slope=-12070, Ea=100kJ/mol, k= 0.000717(45C), 0.00284(55C), 0.00492(65C), 0.0165(75C), 0.0396(85C)
According to Arrhenius equation:
i.e. ln(k2/k1) = -Ea/R (1/T2 - 1/T1)
Where, k1 = 0.000717, T1 = 45 oC = (45+273) K = 318 K
T2 = 25 oC = (25 + 273) K = 298 K
i.e. ln(k2/0.000717) = -12070 (1/298 - 1/318)
i.e. ln(k2/0.000717) = -2.54738
i.e. k2/0.000717 = e-2.54738 = 0.078286
Therefore, the required constant (k2) = 0.078286 * 0.000717 = 5.61*10-5
Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation...
Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calculate the value of the rate constant at 25 °C. Alternatively, you can simply extrapolate the straight line plot of ln(k) vs. 1/T in your notebook to 1/298 , read off the value of ln(k), and determine the value of k. Please put your answer in scientific notation. slope=-12070, Ea=100kJ/mol, k= 0.000717(45C), 0.00284(55C), 0.00492(65C), 0.0165(75C), 0.0396(85C)
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...
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...
QUESTION 3 Assume that, in part A, you determined the slope of your Arrhenius plot to be 6,662 K. Calculate the corresponding activation energy (Ea) in kj/mol corresponding to this value. Report your answer to three (3) significant figures. 1 points Save Answer
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...