The equilibrium constant Kc at 298 K for the reaction A + B ⇌ C + D is 100. Starting with an equimol — Chemical Equilibrium Chemistry Question
Question
The equilibrium constant Kc at 298 K for the reaction A + B ⇌ C + D is 100. Starting with an equimolar solution with concentrations of A, B, C and D all equal to 1 M, the equilibrium concentration of D is_____ × 10 M. (Nearest integer) –2
💡 Solution & Explanation
**Step 1: Set up the ICE table** Reaction: A + B ⇌ C + D | | A | B | C | D | |---|---|---|---|---| | Initial | 1 | 1 | 1 | 1 | | Change | –x | –x | +x | +x | | Equilibrium | 1–x | 1–x | 1+x | 1+x | **Step 2: Apply the equilibrium constant expression** $$K_c = \frac{[C][D]}{[A][B]} = 100$$ $$\frac{(1+x)(1+x)}{(1-x)(1-x)} = 100$$ **Step 3: Simplify the equation** $$\frac{(1+x)^2}{(1-x)^2} = 100$$ Taking the square root of both sides: $$\frac{1+x}{1-x} = 10$$ **Step 4: Solve for x** $$1 + x = 10(1-x)$$ $$1 + x = 10 - 10x$$ $$11x = 9$$ $$x = \frac{9}{11} = 0.8182$$ **Step 5: Calculate equilibrium concentration of D** $$[D]_{eq} = 1 + x = 1 + 0.8182 = 1.8182 \text{ M}$$ **Step 6: Express in the required form** $$1.8182 = 182 × 10^{-2}$$ Therefore, the answer is 182.00.