What is the equilibrium constant for reaction below at 25 °C? (R = 8.314 J/K?mol) 2 NO(g) + O₂(g) $\leftrightharpoons$ 2 NO₂(g); ?_fG° [NO(g)] = +86.6 kJ/mol and ?_fG° [NO₂(g)] = +51.2 kJ/mol. A) 3.9 × 10^-13 B) 1.0 × 10-¹¹ C) 2.6 × 10¹² D) 1.6 × 10⁶ E) 3.8 × 10²⁸

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The equilibrium constant (K) is related to the standard free energy change (?G°) by the equation:K = e^(-?G°/RT)where R is the gas constant (8.314 J/K?mol) and T is the temperature in Kelvin (25 °C = 298 K).Substituting the given values into the equation, we get:K = e^(-(2 * 51.2 kJ/mol - 2 * 86.6 kJ/mol) / (8.314 J/K?mol * 298 K)) = 2.6 × 10^12Therefore, the equilibrium constant for the reaction at 25 °C is 2.6 × 10^12.

What Is the Equilibrium Constant for Reaction Below at 25 ⇋\leftrightharpoons⇋

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