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Calculate the enthalpy of desorption for the process of magnesium oxide (MgO) surface drying at a temperature of 500 K, given that the equilibrium pressure of water vapor above the dried MgO is 0.056 atm and the enthalpy of adsorption of water on MgO is -44.48 kJ/mol.

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To calculate the enthalpy of desorption, we can use the Clausius-Clapeyron equation, which relates the equilibrium pressure of a substance with its temperature and enthalpy of phase transition  in this case, desorption . The equation is:ln P2/P1  = -H/R *  1/T2 - 1/T1 In this case, we are given the equilibrium pressure  P2  of water vapor above the dried MgO at 500 K  T2  and the enthalpy of adsorption of water on MgO. The enthalpy of desorption is the opposite of the enthalpy of adsorption, so:H_desorption = -H_adsorptionH_desorption = - -44.48 kJ/mol  = 44.48 kJ/molSince we are only given the equilibrium pressure at 500 K, we cannot directly use the Clausius-Clapeyron equation to find the enthalpy of desorption. However, we have already determined that the enthalpy of desorption is equal to the enthalpy of adsorption but with the opposite sign. Therefore, the enthalpy of desorption for the process of magnesium oxide surface drying at 500 K is 44.48 kJ/mol.

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