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Calculate the enthalpy of adsorption for the adsorption of nitrogen gas on activated carbon at 298 K, given that the partial pressure of nitrogen gas is 1.2 atm and the amount of adsorbate adsorbed per gram of adsorbent is 0.45 mol/g. The enthalpy of liquefaction of nitrogen gas at boiling point (-196 °C) is -5.58 kJ/mol.

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To calculate the enthalpy of adsorption, we can use the Clausius-Clapeyron equation, which relates the change in vapor pressure with temperature for a phase transition. In this case, we will consider the adsorption of nitrogen gas onto activated carbon as a phase transition from gas to adsorbed state.The Clausius-Clapeyron equation is given by:ln P2/P1  = -H/R *  1/T2 - 1/T1 where P1 and P2 are the initial and final partial pressures of nitrogen gas, H is the enthalpy of adsorption, R is the gas constant  8.314 J/molK , and T1 and T2 are the initial and final temperatures.We are given the partial pressure of nitrogen gas  P2  as 1.2 atm. Since the gas is initially in the atmosphere, we can assume the initial partial pressure  P1  to be 1 atm. The initial temperature  T1  is the boiling point of nitrogen, which is -196 C or 77.15 K. The final temperature  T2  is 298 K.We can rearrange the equation to solve for H:H = -R *  1/T2 - 1/T1  / ln P2/P1 Plugging in the given values:H = -8.314 J/molK *  1/298 K - 1/77.15 K  / ln 1.2 atm/1 atm H = -8.314 J/molK *  0.00336 K^-1 - 0.01296 K^-1  / ln 1.2 H = -8.314 J/molK *  -0.0096 K^-1  / 0.1823H = 443.7 J/molNow, we need to consider the enthalpy of liquefaction of nitrogen gas, which is -5.58 kJ/mol or -5580 J/mol. The enthalpy of adsorption includes both the enthalpy of liquefaction and the enthalpy change associated with the adsorption process. Therefore, we can calculate the total enthalpy of adsorption as:H_adsorption = H + H_liquefactionH_adsorption = 443.7 J/mol - 5580 J/molH_adsorption = -5136.3 J/molThe enthalpy of adsorption for the adsorption of nitrogen gas on activated carbon at 298 K is approximately -5136.3 J/mol.

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