Question

Calculate the enthalpy of adsorption for a reaction where 2 moles of nitrogen gas (N2) are adsorbed on the surface of a solid at -100°C. Given that the partial pressure of N2 in the gas phase is 0.2 atm and the equilibrium constant for adsorption of N2 at this temperature is 2.5 x 10^3. Assume the adsorption is a reversible process and that the ideal gas law can be used to calculate partial pressures.

Answer 1

To calculate the enthalpy of adsorption, we can use the van't Hoff equation, which relates the equilibrium constant  K  to the enthalpy change  H  and temperature  T :ln K  = -H / R *  1/T  + Cwhere R is the gas constant  8.314 J/mol*K  and C is a constant.We are given the equilibrium constant  K  for the adsorption of N2 at -100C  173.15 K  and the partial pressure of N2 in the gas phase  0.2 atm . We can rearrange the van't Hoff equation to solve for H:H = -R * T * ln K  /  1/T First, we need to convert the partial pressure of N2 from atm to Pa:0.2 atm *  101325 Pa/atm  = 20265 PaNow, we can plug in the values into the van't Hoff equation:H = -  8.314 J/mol*K  *  173.15 K  * ln 2.5 x 10^3  /  1/173.15 K H = -  8.314 J/mol*K  *  173.15 K  * ln 2.5 x 10^3  *  173.15 K H  -  8.314 J/mol*K  *  173.15 K  * 7.824H  -11237.4 J/molSince the adsorption process involves 2 moles of N2, we need to divide the enthalpy change by 2 to obtain the enthalpy of adsorption per mole of N2:H_adsorption = -11237.4 J/mol / 2H_adsorption  -5618.7 J/molThe enthalpy of adsorption for the reaction is approximately -5618.7 J/mol.