2. (5 points) Consider the oxidation reaction where methane, a powerful green-ho
ID: 711316 • Letter: 2
Question
2. (5 points) Consider the oxidation reaction where methane, a powerful green-house gas, is catalytically oxidized to CO2 and H20 by Pdr-Al O3 catalyst CH4 + 202 H+CO2 How would you determine the activation energy where you adjust (very high SV) so the % conversion for each reactant is less than 5% (differential reaction conditions) for the following three measured relative rates? Draw the best straight line through the points and calculate the average activation energy. Rate (312 K.) 29.2 Rate (297 °C) = 78.3 Rate (282cc) 58.5Explanation / Answer
Q1: How would you determine the Activation energy for the following measured relative rates?
A series of experimets are conducted at differntial reactor conditions where the concentrations are kept constant and temperature is varied. Once rates are obtained for different temperatures as shown in context, a simple plot of natural log of realtive rates versus inverse of temperature can be drawn. One can find out the activation energy, by calculating the slope of the plot which would equal -Ea/R.
Q2: Draw the best straight line and calculate the average Activation energy.
Using the arrehineous equation:
K = Ae-Ea/(RT) where K is the realtive rate, A is the pre exponential factor, Ea is the activation energy, R is the universal gas constant, and T is the temperature.
Applying natural log (ln) on both sides and rearraging the equation, we get:
ln(k) = (- Ea/R)(1/T) + lnA;
So a plot of ln(k) vs 1/T gives a straight line with slope equivalent to -Ea/R, with a "y" intercept of lnA. Since the value of R is known, one can find out the activation energy by calculating the slope of the line.
Aim: to calculate slope of the line using given data.
Given, Relative Rate (K1) = 99.9, we need to calculate ln(K1) = ln(99.9) = 4.60416969
Similarly calculate:
Relative Rate (K2) = 78.3, therefore ln(K2) = 4.3605476
Relative Rate (K3) = 58.5, therefore ln(K3) = 4.0690268
Given, T1 = 3120C.
Convert T1 in to Kelvin, T1= 312+273.15 K = 585.15 K and calculate 1/T1 = 1/(585.15) = 0.00170896 K-1
Similarly T2 in Kelvin = 297+273.15 K = 570.15 K and calculate 1/T2 = 1/(570.15) = 0.00175392
and T3 in kelvin = 282 + 273.15 K = 555.15K and 1/T3 =1/(555.15) = 0.00180131
Now, plot a graph of ln(k) vs 1/T, using the data from the above table:
The slope of the above line calculated = -5797.85
-Ea/R = -5797.85
The average activation energy Ea = 5797.85 Kx 8.314 J/mol.K = 48203.32 J/mol = 48.2 KJ/mol.
K ln(K) T(0C) T(K) 1/T 99.9 4.60416969 312 585.15 0.000170896 78.3 4.3605476 297 570.15 0.00175392 58.5 4.06902675 282 555.15 0.00180131Related Questions
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