Examine the graph below and determine the value of the energy of activation for the chemical...
Examine the graph below and determine the value of the energy of activation for the chemical process. 0 y=-11624x + 32.055 R2 = 1 2 In k -3 -5 -6 0.0028 0.0029 0.0032 0.0033 0.003 0.0031 1/T (1/K) -96654 J/mol +96654 J/mol - 11624 J/mol +11624 J/mol -32.055 J/mol +32.055 J/mol
Examine the graph below and determine the value of the energy of activation for the chemical process. 0 -1 y = -11624x + 32.055 R2 = 1 -2 In k 3 -4 -5 -6 0.0028 0.0029 0.003 0.0031 0.0032 0.0033 1/T (1/K) -32.055 J/mol +32.055 J/mol -96654 J/mol +11624 J/mol - 11624 J/mol +96654 J/mol
Examine the graph below and determine the value of the energy of activation for the chemical process -1 y=-11624x + 32.055 R=1 Ink - 1/T (1/6) +11624 J/mol -32.055 J/mol - 11624 J/mol +32.055 J/mol -96654 J/mol +96654 J/mol
Question 4 2 pts You generate the below Arrhenius plot of your data. Determine the activation energy of the reaction in kJ/mol. Note: R = 8.314 J/(mol*K) = 0.008314 kJ/(mol*K) = 0.0821 L*atm/(mol*K) Arrhenius Plot: Sample Data o e đ d c In(k) đ ở s y = -18808x + 56.183 R2 = 0.976 do o o 0.0032 0.00325 0.0033 0.00335 0.0034 0.00345 1/T
The rate coefficient of a chemical reaction is studied as a function of temperature, and the In kis plotted as a function of 1/T below. What is the activation energy of the chemical reaction in kJ mol-1 5.0- 3.0 Ink 2.04 1.0 0.0031 0.0032 0.0033 0.0034 0.0035 1/T(K) Give your answer correct to two decimal places and do not include the correct units (kJ mol-?) in your answer.
A series of experiments were conducted where the equilibrium constant (K) was determined at various temperatures (T). The data was then graphed (shown below). The equation of the line is listed on the graph. Use the graph to determine AS (in J/mol·K). -30 0.0031 -30.30.003 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037 -31 -31.5 y = -6861.2x - 9.2495 In (K) -32 -32.5 -33 -33.5 -34 -34.5 -35 1/T in K. Answer:
A series of experiments were conducted where the equilibrium constant (K) was determined at various temperatures (T). The data was then graphed (shown below). The equation of the line is listed on the graph. Use the graph to determine AH° (in kJ/mol). -30 0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037 -30.50.003 -31 y = -6861.2x - 9.2495 -31.5 -32 -32.5 In -33 -33.5 -34 -34.5 -35 1/T in K
The graph below can be used to find the activation energy, E_a, for a chemical reaction. Label the Y-axis on the graph. 4.B.3.c. Calculate the activation energy for the reaction in question.
Question 24 (1 point) The graph below shows the results of an Arrhenius plot for the reaction 2 N2O(g) --> 4 NO2(g) + O2(g) In(k) vs. 1/1 In(k) (k in units of s.1) slope-1.24x104 K intercept=315 -14 0.0026 0.0028 0.003 0.0032 0.0034 Т (1/к) Which of the following most correctly describes the activation energy and frequency factor for this reaction? Ea-1.03x102 kJ mol 1 A-3.30x1031 - 1 Ea=-1.24x104 kJ mol-1, A=31.5 s -1 O Ea-1.03x108 kJ mol-1 A-4.79x1013 5-1 Ea=1.03x102...
The rate data of the reaction was collected at different temperatures (°C) and functions of the data plotted to determine the activation energy. 18 17 T 16.5 36.2 47.7 57.3 k 9.86 x 104 2.01 x 106 9.84 x 106 3.41 x 107 16 y = (-1.37 x 10^)x + 58.8 15 14 13 12 = 11 10 0.0030 0.0031 0.0032 0.0033 0.0034 0.0035 1. Clearly label the x and y axes (above) with the proper function. 2. Determine Ea...