36. Solution to Chemistry Problem
📂 Chemistry
👤 Derkach
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1. Using the data tables provided for each option, plot three isotherms of benzene adsorption by graphitic soot.
2. Using the linear form of the Langmuir equation, plot p/A versus p. Determine the limit of monomolecular adsorption (A) and the value of K for three temperatures using the slope of the angle to the abscissa axis and the length of the image along the ordinate axis.
3. Calculate the specific area of the adsorbent using the value of maximum monomolecular adsorption (A) at a temperature of T = 293 K (the area of a benzene molecule on the surface of graphitized soot is 49.10 * 10-20 m2) /molecule.
4. Plot a graph of lnk = f (1/T) and determine the value of ΔH.
5. A = 0.01 mol / kg; A = 0.015 mol / kg; Plot isosteres for A = 0.02 mol/kg and calculate the differential isosteric heat of adsorption from the graphs.
b. Analyze the results.
Option 5
Pa 0 14 34 54 74
A293, mol/kg 0 0.0115 0.0222 0.0293 0.0343
A303, mol/kg 0 0.0070 0.0146 0.0205 0.0250
A313, mol/kg 0 0.0042 0.0094 0.0137 0.0174
2. Using the linear form of the Langmuir equation, plot p/A versus p. Determine the limit of monomolecular adsorption (A) and the value of K for three temperatures using the slope of the angle to the abscissa axis and the length of the image along the ordinate axis.
3. Calculate the specific area of the adsorbent using the value of maximum monomolecular adsorption (A) at a temperature of T = 293 K (the area of a benzene molecule on the surface of graphitized soot is 49.10 * 10-20 m2) /molecule.
4. Plot a graph of lnk = f (1/T) and determine the value of ΔH.
5. A = 0.01 mol / kg; A = 0.015 mol / kg; Plot isosteres for A = 0.02 mol/kg and calculate the differential isosteric heat of adsorption from the graphs.
b. Analyze the results.
Option 5
Pa 0 14 34 54 74
A293, mol/kg 0 0.0115 0.0222 0.0293 0.0343
A303, mol/kg 0 0.0070 0.0146 0.0205 0.0250
A313, mol/kg 0 0.0042 0.0094 0.0137 0.0174
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