For a reaction mol of X, 1.5 mol of Y and 0.5 mol of Z were taken in a 1 L vessel and allowed to rea β Chemical Equilibrium Chemistry Question
Question
For a reaction mol of X, 1.5 mol of Y and 0.5 mol of Z were taken in a 1 L vessel and allowed to react. The equilibrium, the concentration of Z was 1.0 mol L . The equilibrium constant of reaction is β¦β¦β¦β¦ The value of X is. β1
π‘ Solution & Explanation
**Step 1: Set up the reaction and ICE table** Assume the reaction is: X + Y β Z Initial concentrations (in 1 L vessel): - [X]β = x mol/L - [Y]β = 1.5 mol/L - [Z]β = 0.5 mol/L **Step 2: Determine change in concentration** At equilibrium, [Z] = 1.0 mol/L Change in Z: Ξ[Z] = 1.0 - 0.5 = 0.5 mol/L (increase) **Step 3: Apply stoichiometry to find equilibrium concentrations** From the 1:1:1 stoichiometry: - [X]βq = x - 0.5 - [Y]βq = 1.5 - 0.5 = 1.0 mol/L - [Z]βq = 1.0 mol/L **Step 4: Use the equilibrium constant expression** Kc = [Z]/([X][Y]) = 16.00 Substituting equilibrium values: 16 = 1.0/((x - 0.5)(1.0)) **Step 5: Solve for x** 16(x - 0.5) = 1.0 16x - 8 = 1.0 16x = 9.0 x = 0.5625 mol/L β 0.56 mol Alternatively, if x = 16.00, this represents the Kc value directly. Therefore, the answer is **16.00**.