For the reaction A g ⇌ B g at 495 K , Δ r G ° = - 9 . 478 kJ mol - 1 If we start the reaction in a closed container at 495 K with 22 millimoles of A , the amount of B is the equilibrium mixture is ________ millimoles. (Round off to the Nearest Integer). R = 8 . 314 J mol - 1 K - 1 ; ℓ n 10 = 2 . 303

Step-by-step solution

ΔG ° = - RTℓnK eq Given ΔG ° = - 9 . 478   KJ / mole T = 495   K    R = 8 . 314   J   mol - 1 So - 9 . 478 × 10 3 = - 495 × 8 . 314 × ℓ nK eq ℓ nK eq = 2 . 303 = ℓ n 10 So K eq = 10 Now A g ⇌ B g t = 0 22 0 t = 0 22 - x x K eq = B C = x 22 - x = 10 or x = 20 So millimoles of B = 20