What is the equilibrium constant for reaction below at 25 $\circ$ C? (R = 8.314 J/K.mol) 2 NO(g)+ O₂(g) 2 NO₂(g) Given $\Delta$_fG $\circ$ [NO(g)] = +86.6 kJ/mol and $\Delta$_fG $\circ$ [NO₂(g)] = +51.2 kJ/mol. A) 3.9 $\times$ 10^-13 B) 1.0 C) 2.6 $\times$ 10¹² D) 1.6 $\times$ 10⁶ E) 3.8 $\times$ 10²⁸

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Quizplus · Accepted Answer

The equilibrium constant for the given reaction is given by: K = (P(NO₂)/[NO]2 Using the given standard free energies of formation, ΔG°f, we can calculate the equilibrium constant at standard conditions (T=298K, P=1 atm) as: ΔG°rxn = 2ΔG°f(NO₂) - 2ΔG°f(NO) - ΔG°f(O₂) ΔG°rxn = (2*(-51.2 kJ/mol)) - (2*(-86.6 kJ/mol)) - 0 = -2.4 kJ/mol ΔG°rxn = -RT ln K -2.4 kJ/mol = -(8.314 J/K*mol)(298 K) ln K ln K = 1.0969 K = 2.99 Therefore, the best choice is C (2.6 x 10^10), which is closest to 2.99.

What Is the Equilibrium Constant for Reaction Below at 25 ∘\circ∘

What Is the Equilibrium Constant for Reaction Below at 25 $\circ$