Question

Calculate the enthalpy of adsorption when 2.5 grams of hydrogen gas is adsorbed onto a surface at a temperature of 298 K. The equilibrium pressure of hydrogen gas at this temperature is 1.5 atm and the surface area of the adsorbent is 50 cm². (Assume ideal gas behavior and that the adsorption occurs under standard conditions.)

Answer 1

To calculate the enthalpy of adsorption, we can use the Clausius-Clapeyron equation:H_ads = -RT * ln P/P where H_ads is the enthalpy of adsorption, R is the ideal gas constant  8.314 J/molK , T is the temperature in Kelvin  298 K , P is the equilibrium pressure  1.5 atm , and P is the standard pressure  1 atm .First, we need to convert the pressure from atm to Pa:1 atm = 101325 PaP = 1.5 atm * 101325 Pa/atm = 151987.5 PaP = 1 atm * 101325 Pa/atm = 101325 PaNow, we can plug the values into the Clausius-Clapeyron equation:H_ads = -8.314 J/molK * 298 K * ln 151987.5 Pa / 101325 Pa H_ads = -8.314 J/molK * 298 K * ln 1.5 H_ads  -3486.9 J/molNow, we need to find the number of moles of hydrogen gas:Molar mass of H = 2 g/molNumber of moles = mass / molar mass = 2.5 g / 2 g/mol = 1.25 molFinally, we can calculate the total enthalpy of adsorption:Total enthalpy of adsorption = H_ads * number of molesTotal enthalpy of adsorption = -3486.9 J/mol * 1.25 molTotal enthalpy of adsorption  -4358.6 JSo, the enthalpy of adsorption for 2.5 grams of hydrogen gas adsorbed onto the surface is approximately -4358.6 J.