The molar enthalpy of formation of water vapor is -241.82 kJ/mol at298 K. Calcul

Question

The molar enthalpy of formation of water vapor is -241.82 kJ/mol at298 K. Calculate the molar enthalpies of formation of water vaporand of liquid water at 100°C and at -0.1°C (fourvalues)

Explanation / Answer

298 K is 25 C H2(g) + 0.5*O2 (g) = > H2O(g)       H1 = -241.82kJ/mol at 100 C, H = Cp*T +H1,       where Cp =[Cp(H2O(g)) - Cp(H2) - 0.5*Cp(O2)] T = (100 C - 25 C) = 75 C H(form water gas) = -241.82 kJ/mol + (75 C)*(37.47 J/mol/C -28.836 J/mol/C - 0.5*29.378 J/mol/C)*1 kJ/1000J H(form water gas) = -242.274125 kJ/mol ~ -242.3kJ/mol   (at 100 C) At -0.1 C, H(form water gas) = -241.82 kJ/mol + (-0.1 - 25)C*(37.47 J/mol/C - 28.836 J/mol/C -0.5*29.378 J/mol/C)*1 kJ/1000J = -241.668019 ~ -241.67 kJ/mol =================================== formation of water H2O(g) =>   H2O (l)       H2 = -40.65 at 25C Add to H2(g) + 0.5*O2 (g) = > H2O(g)       H1 = -241.82kJ/mol So that H2(g) + 0.5*O2 (g) = > H2O(l)     H = H1 + H2, but H2 needs to be corrected for temperature H2 (T) = H2(T = 25 C) + [Cp(H2O(l)) -Cp(H2O(g))]*T H2(100C) = -40.65 kJ/mol + (74.539 J/mol/C - 37.47 J/mol/C)*(100C - 25 C) *(1 kJ/1000J) H2(100C) = -37.869825 kJ/mol H2(-0.1 C) = -40.65 kJ/mol + (74.539 J/mol/C - 37.47 J/mol/C)*(-0.1 - 25 C) *(1 kJ/1000J) H2(-0.1 C) = -41.5804319 kJ/mol H (100) = H1 (100 C) + H2(100C) = -242.274125kJ/mol + -37.869825 kJ/mol H(T = 100 C) = -280.14395 kJ/mol ~ -280.14 kJ/mol H(-0.1 C) = H1 ( -0.1 C) = H2(-0.1 C) =-241.668019 kJ/mol + -41.5804319 kJ/mol H(-0.1 C) = -283.248451 ~ -283.25 kJ/mol

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