Consider the synthesis of ammonia by the Haber process: N ₂ (g) + 3 H ₂ (g) 2 NH ₃ (g) At 300 K, the standard internal energy change ( U^ ) for the reaction is -87.2 kJ mol ⁻¹ and the standard entropy change ( S^ ) is -198 J K ⁻¹ mol ⁻¹ . Identify the correct option with the standard Gibbs free energy change ( G^ ) and the spontaneity of the reaction at 300 K. (Given: R = 8.3 J K ⁻¹ mol ⁻¹ )

Step-by-step solution

The change in the number of moles of gaseous species is: n_g = n_p - n_r = 2 - (1 + 3) = -2 The standard enthalpy change H^ is given by: H^ = U^ + n_g R T Substituting the given values (converting R to kJ K ⁻¹ mol ⁻¹ ): H^ = -87.2 + (-2) (8.3 10⁻³) 300 H^ = -87.2 - 4.98 = -92.18 kJ mol ⁻¹ The standard Gibbs free energy change G^ is: G^ = H^ - T S^ G^ = -92.18 - 300 (-198 10⁻³) G^ = -92.18 + 59.4 = -32.78 kJ mol ⁻¹ Since G^ Using U^ directly in place of H^ leads to the incorrect value of -27.80 kJ mol ⁻¹ . Answer: -