A chemistry student needs to calculate the enthalpy of adsorption of ammonia that is adsorbed on silica gel at a temperature of 298 K. The student is provided with the following information: - The partial pressure of ammonia in the gas phase is 0.5 atm- The initial temperature of the clean silica gel is 298 K- The final temperature of the adsorbed ammonia and silica gel mixture is 306 K- The mass of ammonia adsorbed on silica gel is 0.025 grams What is the enthalpy of adsorption of ammonia on silica gel at 298 K, in kJ/mol?

Question

A chemistry student needs to calculate the enthalpy of adsorption of ammonia that is adsorbed on silica gel at a temperature of 298 K. The student is provided with the following information: - The partial pressure of ammonia in the gas phase is 0.5 atm- The initial t

Answer 1

To calculate the enthalpy of adsorption, we first need to determine the amount of heat released during the adsorption process. We can do this using the heat capacity of ammonia and the temperature change of the system.1. Calculate the moles of ammonia adsorbed:Ammonia has a molar mass of 17 g/mol. moles of ammonia = mass of ammonia / molar massmoles of ammonia = 0.025 g / 17 g/mol = 0.00147 mol2. Calculate the heat released during adsorption:We will use the heat capacity of ammonia, which is approximately 35 J/molK.T = T_final - T_initial = 306 K - 298 K = 8 Kq = moles of ammonia  heat capacity  Tq = 0.00147 mol  35 J/molK  8 K = 0.411 J3. Calculate the enthalpy of adsorption per mole of ammonia:H_adsorption = q / moles of ammoniaH_adsorption = 0.411 J / 0.00147 mol = 279.6 J/mol4. Convert the enthalpy of adsorption to kJ/mol:H_adsorption = 279.6 J/mol   1 kJ / 1000 J  = 0.280 kJ/molThe enthalpy of adsorption of ammonia on silica gel at 298 K is approximately 0.280 kJ/mol.