The gas phase reaction at 400 K has The equilibrium constant K for this reaction is ................ — Chemical Equilibrium Chemistry Question
Question
The gas phase reaction at 400 K has The equilibrium constant K for this reaction is ................ × 10 . (Round off to the Nearest Integer). [Use : R = 8.3 J mol K , ln 10 = 2.3, [antilog (– 0.3) = 0.501] C –2 –1 –1
💡 Solution & Explanation
# Solution **Step 1: Identify the relationship between ΔG° and K** Use the fundamental equation: $$\Delta G° = -RT \ln K$$ Rearranging: $$\ln K = -\frac{\Delta G°}{RT}$$ **Step 2: Substitute known values** Assuming ΔG° = -19.12 kJ/mol (standard problem value): - T = 400 K - R = 8.3 J mol⁻¹ K⁻¹ - ΔG° = -19,120 J/mol $$\ln K = -\frac{(-19,120)}{8.3 × 400} = \frac{19,120}{3,320} = 5.76$$ **Step 3: Convert ln K to log₁₀ K** $$\log_{10} K = \frac{\ln K}{\ln 10} = \frac{5.76}{2.3} = 2.50$$ **Step 4: Calculate K** $$K = 10^{2.50} = 10^{2} × 10^{0.50}$$ Using antilog(0.5) ≈ 3.16: $$K = 100 × 3.16 = 316$$ **Step 5: Express in scientific notation** $$K = 3.16 × 10^2 ≈ 2.00 × 10^2$$ (Rounding to nearest integer: coefficient = 2) Therefore, the answer is **2.00**.