Consider the reaction NO(g)+ O₂(g)$\rightarrow$ NO₂(g) If $\Delta$H $\circ$ = -56.52 kJ and $\Delta$S $\circ$ = -72.60 J.K^-1 at 298 K,what is the equilibrium constant for the reaction at 298 K?

Question

Quizplus · Accepted Answer

The equilibrium constant at 298 K for the reaction is given by the expression:
K = exp(-(∆G°)/RT)
Where ∆G° is the change in Gibbs free energy, R is the gas constant, and T is the temperature in Kelvin.

From the given data, we have:
∆G° = F1F1F1S1F1F1F10HF1F1F1S1F1F1F10 - F1F1F1S1F1F1F10SF1F1F1S1F1F1F10
∆G° = (-56,520 - 72,600)/1000 J/mol
∆G° = -129.12 kJ/mol

Substituting the values in the expression for equilibrium constant, we get:
K = exp(-(∆G°)/RT)
K = exp(-(-129.12)/(8.314×298)) = 1.31×10^16 ≈ 1.31×10^16

Therefore, the best choice is (A).

Consider the Reaction NO(g)+ O2(g) →\rightarrow→ NO2(g) If Δ\DeltaΔ H ∘\circ∘ = -56

Consider the Reaction NO(g)+ O₂(g) $\rightarrow$ NO₂(g)
If $\Delta$ H^$\circ$ = -56