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The values for the standard formation enthalpies at 298.15K for CO (g), and H2O(g) are -...

The values for the standard formation enthalpies at 298.15K for CO (g), and H2O(g) are - 110.53 kJ/mol and - 241.82 kJ/mol, respectively. The molar heat capacities of products and reactants are given as: CPm Ø (H2(g)) = 28.824 J/K.mol, CPm Ø (CO(g)) = 29.140 J/mol.K, CPm Ø (H2O(g)) = 33.580 J/mol.K, CPm Ø (C(s, graphite)) = 8.527 J/mol.K Calculate Delta (reaction) U in kJ for the formation at 378.15 K and 1 bar of 6.54 g CO(g) following the reaction: C(s, graphite) + H2O(g) --> CO(g) + H2(g)

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Answer #1

From the given data,

Standard enthalpy change at 298.15 K,

dH = dH(products) - dH(reactants)

      = (-110.53) - (-241.82)

      = 131.29 kJ/mol

dCp = Cp(products) - Cp(reactants)

        = (29.140 + 28.824) - (33.580 + 8.527)

        = 15.857 J/K.mol

If Change in internal energy = dU

Thus,

dU = dH + CpdT

      = 131.29 + 0.015857(378.15 - 298.15)

      = 132.56 kJ/mol

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