Calculate the standard enthalpy change for the following reaction at 25 °C.

Question

C3H8(g)+5O2(g)-->3CO2(g)+4H2O(g)

Answer 1

QUESTION

Calculate the standard enthalpy change for the following reaction at 25 C⁰?

C₃H_3(g) + 5 O_2(g) =======> 3 CO_2(g) + 4 H₂O_(g)

SOLUTION

Standard enthalpy change = Δ H⁰

Δ H⁰ for C₃H_3(g) = -103.85 KJ/mol

Δ H⁰ for 5 O_2(g) = 0.00 KJ/mol

Δ H⁰ for 3 CO_2(g) = 3 (-393.51) KJ/mol = - 1180.53 KJ/mol

Δ H⁰ for 4 H₂O_(g) = 4 (-241.82) KJ/mol = -967.28 KJ/mol

Δ H⁰ = ∑Δ H⁰_products - ∑Δ H⁰_reactants

Δ H⁰ = [(-1180.53) KJ/mol + [(-967.28) KJ/mol] - [(-103.85) KJ/mol + (0.00) KJ/mol]

Δ H⁰ = (-2147.81) KJ/mol - (-103.85) KJ/mol

Δ H⁰ = -2043.96 KJ/mol

Thus, the standard enthalpy change for the above reaction is Δ H⁰ = -2043.96 KJ/mol.