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Chemistry · Equilibrium involving chemical processes: Law of chemical equilibrium, equilibrium constants (Kp and Kc) and their significance, the significance of Delta G and Delta G° in chemical equilibrium

For } \rightleftharpoons \boldsymbol{X} \boldsymbol{Y}_{(\boldsymbol{g})}+\bolds

For \( \boldsymbol{X} \boldsymbol{Y}_{\mathbf{2}(\boldsymbol{g})} \rightleftharpoons \boldsymbol{X} \boldsymbol{Y}_{(\boldsymbol{g})}+\boldsymbol{Y}_{(\boldsymbol{g})}, \) intially 1 mole each of \( X Y_{2} \) and \( y \) are present in 10L flask at 500mm.lf the equilibrium pressure of \( \mathrm{XY} \) is \( 150 \mathrm{mm}, K_{p} \) is?

  • A. \( 500 \mathrm{mm} \mathrm{mg} \)
  • B. \( \sqrt{500} \mathrm{mm} \) нв
  • C. \( 300 \mathrm{mm} \) Hg
  • D. 600mm Hg

Step-by-step solution

Initial total pressure is 500 mm Hg with 2 moles. Let α be degree of dissociation. At equilibrium: moles: XY2 = 1-α, XY = α, Y = 1+α; total moles = 2+α. Total pressure = 500×(2+α)/2 = 250(2+α). Partial pressure of XY = [α/(2+α)]×250(2+α) = 250α = 150 → α = 0.6. Total pressure = 250×2.6 = 650 mm. Then P_XY2 = (1-α)/(2+α)×650 = (0.4/2.6)×650 = 100 mm, P_Y = (1+α)/(2+α)×650 = (1.6/2.6)×650 = 400 mm. Kp = (P_XY × P_Y)/P_XY2 = (150×400)/100 = 600 mm Hg.
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