At 1990 K and 1 atm pressure, there are equal number of Cl molecules and Cl atoms in the reaction mi — Chemical Equilibrium Chemistry Question
Question
At 1990 K and 1 atm pressure, there are equal number of Cl molecules and Cl atoms in the reaction mixture. The value of K for the reaction under the above conditions is x × 10 . The value of x is _______. (Rounded off to the nearest integer) 2 p -1
💡 Solution & Explanation
**Step 1: Write the equilibrium reaction** Cl₂(g) ⇌ 2Cl(g) **Step 2: Set up the ICE table** Let initial moles of Cl₂ = a - At equilibrium: Cl₂ = (a - x) moles, Cl = 2x moles - Given condition: moles of Cl atoms = moles of Cl₂ molecules - Therefore: 2x = a - x - Solving: 3x = a, so x = a/3 **Step 3: Determine equilibrium composition** - Cl₂ at equilibrium = a - a/3 = 2a/3 - Cl at equilibrium = 2(a/3) = 2a/3 - Total moles = 2a/3 + 2a/3 = 4a/3 **Step 4: Calculate partial pressures** Using mole fractions: - P(Cl₂) = (2a/3)/(4a/3) × 1 = 0.5 atm - P(Cl) = (2a/3)/(4a/3) × 1 = 0.5 atm **Step 5: Calculate Kp** $$K_p = \frac{[P(Cl)]^2}{P(Cl_2)} = \frac{(0.5)^2}{0.5} = \frac{0.25}{0.5} = 0.5 \text{ atm}$$ **Step 6: Express in the required form** K = 0.5 = 5.0 × 10⁻¹ Since K = x × 10⁻¹, then x = 5.00 Therefore, the answer is 5.00.