The thermal decomposition of calcium carbonate is given by the following equation: CaCO ₃ (s) CaO(s) + CO ₂ (g) At 1000 K, the standard Gibbs free energy change ( G^ ) for this reaction is +22.6 kJ mol ⁻¹ and the standard entropy change ( S^ ) is +160 J K ⁻¹ mol ⁻¹ . What is the standard internal energy change ( U^ ) for the reaction at 1000 K? (Given: R = 8.3 J K ⁻¹ mol ⁻¹ )

Step-by-step solution

First, calculate the standard enthalpy change ( H^ ) using the Gibbs free energy equation: G^ = H^ - T S^ H^ = G^ + T S^ Substituting the given values (converting S^ to kJ K ⁻¹ mol ⁻¹ ): H^ = 22.6 + 1000 (160 10⁻³) H^ = 22.6 + 160 = +182.6 kJ mol ⁻¹ Next, determine the change in the number of moles of gaseous species ( n_g ). Only CO ₂ is gaseous: n_g = 1 - 0 = 1 Now, calculate the standard internal energy change ( U^ ) using the relation: H^ = U^ + n_g R T U^ = H^ - n_g R T U^ = 182.6 - (1) (8.3 10⁻³) 1000 U^ = 18