What is the equilibrium constant for reaction below at 25 °C? (R = 8.314 J/K⋅mol) 2 NO(g)+ O₂(g) 2 NO₂(g); Δ_fG° [NO(g)] = +86.6 kJ/mol and Δ_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

Question

Quizplus · Accepted Answer

We can use the formula for calculating the equilibrium constant (K), which relates to ΔG° at standard conditions through the following equation:
ΔG° = -RT ln(K)
where R is the gas constant, T is the temperature in Kelvin, and ln is the natural logarithm.
First, we need to calculate ΔG° for the reaction by using the equation:
ΔG° = ΣΔG°f(products) - ΣΔG°f(reactants)
where ΔG°f is the standard Gibbs free energy of formation at standard conditions.
ΔG°f(NO) = 86.6 kJ/mol
ΔG°f(OS) = 0 (since it is in its standard state)
ΔG°f(NOS) = 51.2 kJ/mol
ΔG° = [(2 × ΔG°f(NOS)) - (2 × ΔG°f(NO)) - (0)] = -96.4 kJ/mol
Now that we have ΔG°, we can calculate K:
K = e^(-ΔG°/RT) = e^(-(-96400J/mol) / (8.314 J/K⋅mol × 298 K)) = 2.6 × 10^12
Therefore, the best choice is C.

What Is the Equilibrium Constant for Reaction Below at 25