Answer to Question #208540 in Physical Chemistry for pesky

Question

Calculate ΔH at 400^oC for the reaction

SO_2(g) + 1/2O_2(g)⟶SO_3(g)

given that the standard heat formation at 25^oC of SO₂ and SO₃ are -70.9 and -94.4 (in kcal mol^-1) repectively and that the heat capacities in JK^-1mol^-1 are:

SO₂: C_p = 10.39 + 2.54 x 10^-3T -1.42 x 10^-5T²

SO₃: C_p = 13.70 + 6.42 x 10^-3T -3.12 x 10^-5T²

O₂: C_p = 7.16 + 1.00 x 10^-3T -0.40 x 10^-5T²

Answer 1

formation looks like this

ΔHrxn=∑(n⋅ΔH⁰f products)−∑(m⋅ΔH⁰f reactants)

n and m are the stoichiometric coefficients (number of moles) of the products and reactants, respectively.

ΔH⁰f SO₂=-296.84 kJ/mol

ΔH⁰f SO₃=-395.7 kJ/mol

ΔH⁰f O₂=0 kJ/mol

However this enthalpy change involves C_p and SHC have been provided.

Therefore, the enthalpy change of the reaction will be

∆H°T = ∆)H°T₁ + $\intop$_T ∆C_p° (T¹)dT^1-

400-25 = 375°

C_p SO₂ + O₂ = 12.93 ×10^-3T -1.42 ×10^-5T² + 20.12 ×10^-3T- 0.4 × 10^-5T²

= 17.55 + 3.4(10^-3 T) - 1.82(10^-5) T²

C_p = a + bT + CT^-2 + dT²

C_p = -3.85 + 3.02(10^-3)T - 1.3(10^-5)T²

C_p = -3.85 + 0.00303T -0.000013T²

= -3.85 + 0.00303(-70.9) -(0.000013(-94.4)

= -3.85 - 0.0215 + 0.00123

C_p = -3.87

∆H = 400 + (-3.87(-395.7 +296.84)

= 400 + -381.195

= 18.805 kJ