Answer to Question #177467 in Chemistry for Gaurav

Question

The enthalpy of formation of NH3(g) as per the following reaction is -46.11 kJ/mol 

at 298 K. NE(g) + 3H2(g) ----> 2NH3(g)

 Calculate the value of enthalpy of formation of NH2(g) at 100°C. The Cp(J/K/mol)

 

values are given as: N2(g) = 29.12; H2(g) = 28.82; NH3(g) = 35.06. Assume the 

given heat capacity values to be temperature independent in the range. 

Answer 1

Solution:

The balanced chemical equation:

N₂(g) + 3H₂(g) → 2NH₃(g), Δ_rH^o = -46.11 kJ/mol

Assume heat capacities are independent of temp T.

Thus,

Δ_rC_p = ∑C_p(products) − ∑C_p(reactants)

Δ_rC_p = 2×C_p(NH₃) - (3×C_p(H₂) + C_p(N₂))

Δ_rC_p = (2 × 35.06) - (3 × 28.82 + 29.12) = -45.46 J/K/mol

The temperature dependence of enthalpy: dH = C_pdT (at constant pressure).

And so for a temperature change from T₁ to T₂:

If C_p is independent of temperature, then:

Δ_rH(T₂) = Δ_rH(T₁) + Δ_rC_p × (T₂ - T₁)

Δ_rC_p = -45.46 J/K/mol

T₁ = 298 K

T₂ = 100°C + 273 = 373 K

Hence,

Δ_rH at 100°C = -46110 J/mol + (-45.46 J/K/mol) × (373 K - 298 K) = -49519.5 J/mol = -49.52 kJ/mol

Δ_rH at 100°C = -49.52 kJ/mol

Answer: Δ_rH at 100°C = -49.52 kJ/mol.