For the reaction A → B , the value of the equilibrium constant at 300 K and 1 atm is equal to 100.0. — Chemical Equilibrium Chemistry Question
Question
For the reaction A → B , the value of the equilibrium constant at 300 K and 1 atm is equal to 100.0. The value of ∆G for the reaction at 300 K and 1 atm in J mol is –xR, where x is _________. (Rounded off to the nearest integer) [R = 8.31 J mol K and ln10 = 2.3) (g) (g) r 0 -1 -1 -1
💡 Solution & Explanation
**Step 1: Identify the relationship between ΔG° and K** Use the fundamental equation: $$\Delta G° = -RT \ln K$$ **Step 2: Substitute known values** Given: - K = 100.0 - T = 300 K - R = 8.31 J mol⁻¹ K⁻¹ - ln 10 = 2.3 $$\Delta G° = -(8.31)(300) \ln(100)$$ **Step 3: Calculate ln(100)** $$\ln(100) = \ln(10^2) = 2 \ln(10) = 2 × 2.3 = 4.6$$ **Step 4: Calculate ΔG°** $$\Delta G° = -(8.31)(300)(4.6)$$ $$\Delta G° = -2493 × 4.6$$ $$\Delta G° = -11,467.8 \text{ J/mol}$$ **Step 5: Express ΔG° in the form –xR** $$-11,467.8 = -x(8.31)$$ $$x = \frac{11,467.8}{8.31}$$ $$x = 1,380.02$$ **Step 6: Round to nearest integer** $$x ≈ 1,380$$ Therefore, the answer is 1380.00.