Consider the following reaction : 2 A (g) + B (g) 2 D (g) U^ = -10 kJ mol ⁻¹ and S^ = -44 J K ⁻¹ at 298 K. Identify the correct option with G^ for the reaction and spontaneity of the reaction at 298 K. (Given : R = 8.31 J mol ⁻¹ K ⁻¹ )

Step-by-step solution

The change in the number of moles of gaseous species is: n_g = n_p - n_r = 2 - (2 + 1) = -1 The standard enthalpy change H^ is given by: H^ = U^ + n_g R T Substituting the given values: H^ = -10 + (-1) (8.31 10⁻³) 298 H^ = -10 - 2.47638 = -12.47638 kJ mol ⁻¹ The standard Gibbs free energy change G^ is: G^ = H^ - T S^ G^ = -12.47638 - 298 (-44 10⁻³) G^ = -12.47638 + 13.112 = +0.63562 kJ mol ⁻¹ Since G^ > 0 , the reaction is non-spontaneous. Answer: +0.63568 kJ mol ⁻¹ , non-spontaneous