This problem requires calculating equilibrium concentrations for the reaction H₂(g) + I₂(g) ⇌ 2HI(g). With initial concentrations [H₂] = 0.80 M, [I₂] = 0.80 M, and [HI] = 8.00 M, we calculate Q = [HI]²/([H₂][I₂]) = (8.00)²/(0.80)(0.80) = 100. Since Q > Kc (100 > 64), the reaction shifts left. Using an ICE table where [HI] changes by -2x: [H₂] = 0.80 + x, [I₂] = 0.80 + x, and [HI] = 8.00 - 2x at equilibrium. Substituting into Kc = 64: (8.00 - 2x)²/((0.80 + x)²) = 64, taking the square root gives (8.00 - 2x)/(0.80 + x) = 8. Solving: 8.00 - 2x = 6.40 + 8x, which yields x = 0.25, so [HI]eq = 8.00 - 2(0.25) = 7.50 M. The distractor 8.00 M (choice D) incorrectly assumes the system is already at equilibrium. The strategy involves calculating Q, determining shift direction, and solving the equilibrium expression with proper stoichiometry.

This problem requires calculating equilibrium concentrations using an ICE table. For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), we set up the ICE table with initial concentrations [H₂] = 1.00 M, [I₂] = 1.00 M, and [HI] = 10.0 M. Since we're told only [HI] changes appreciably, we let the change in [HI] be -2x (negative because Q > K, so the reaction shifts left), making changes in [H₂] and [I₂] each +x. At equilibrium: [H₂] = 1.00 + x, [I₂] = 1.00 + x, and [HI] = 10.0 - 2x. Substituting into Kc = [HI]²/([H₂][I₂]) = 100 gives (10.0 - 2x)²/((1.00 + x)(1.00 + x)) = 100, which simplifies to (10.0 - 2x)² = 100(1.00 + x)². Taking the square root: 10.0 - 2x = 10(1.00 + x), solving gives x = 0.25, so [HI]eq = 10.0 - 2(0.25) = 9.50 M. A common error (choice B, 10.0 M) assumes no change occurs because the initial [HI] seems large. The key strategy is to properly set up the ICE table with correct stoichiometric coefficients and solve the resulting equation systematically.

This problem requires calculating equilibrium concentrations for H₂(g) + I₂(g) ⇌ 2HI(g). With initial concentrations [H₂] = 0.70 M, [I₂] = 0.70 M, and [HI] = 7.00 M, we calculate Q = [HI]²/([H₂][I₂]) = (7.00)²/(0.70)(0.70) = 100. Since Q > Kc (100 > 49), the reaction shifts left to reach equilibrium. Setting up the ICE table with [HI] decreasing by 2x: [H₂] = 0.70 + x, [I₂] = 0.70 + x, and [HI] = 7.00 - 2x at equilibrium. Substituting into Kc = 49: (7.00 - 2x)²/((0.70 + x)²) = 49, taking the square root gives (7.00 - 2x)/(0.70 + x) = 7. Solving: 7.00 - 2x = 4.90 + 7x, which gives x = 0.10, so [HI]eq = 7.00 - 2(0.10) = 6.80 M. The distractor 7.00 M (choice B) incorrectly assumes the system is already at equilibrium. The key strategy is to always check Q versus K and solve systematically using the equilibrium expression.

This problem involves calculating equilibrium concentrations for H₂(g) + I₂(g) ⇌ 2HI(g). Setting up the ICE table with initial values [H₂] = 0.50 M, [I₂] = 0.50 M, and [HI] = 5.00 M, we need to determine which direction the reaction shifts. Calculating Q = [HI]²/([H₂][I₂]) = (5.00)²/(0.50)(0.50) = 100, and since Q > Kc (100 > 25), the reaction shifts left. With [HI] changing by -2x, we have [H₂] = 0.50 + x, [I₂] = 0.50 + x, and [HI] = 5.00 - 2x at equilibrium. Substituting into Kc = 25: (5.00 - 2x)²/((0.50 + x)²) = 25, taking the square root gives (5.00 - 2x)/(0.50 + x) = 5. Solving: 5.00 - 2x = 2.50 + 5x, which gives x = 0.25, so [HI]eq = 5.00 - 2(0.25) = 4.50 M. A tempting distractor (choice D, 5.00 M) incorrectly assumes the system is already at equilibrium. The strategy is to always calculate Q first, compare to K to determine shift direction, then solve the equilibrium expression.

This question involves calculating equilibrium concentrations for H₂(g) + I₂(g) ⇌ 2HI(g). Starting with [H₂] = 0.30 M, [I₂] = 0.30 M, and [HI] = 6.00 M, we calculate Q = [HI]²/([H₂][I₂]) = (6.00)²/(0.30)(0.30) = 400. Since Q > Kc (400 > 36), the reaction must shift left to reach equilibrium. Setting up the ICE table with [HI] decreasing by 2x: [H₂] = 0.30 + x, [I₂] = 0.30 + x, and [HI] = 6.00 - 2x at equilibrium. Substituting into Kc = 36: (6.00 - 2x)²/((0.30 + x)²) = 36, taking the square root gives (6.00 - 2x)/(0.30 + x) = 6. Solving: 6.00 - 2x = 1.80 + 6x, which gives x = 0.25, so [HI]eq = 6.00 - 2(0.25) = 5.50 M. A common error (choice C, 6.00 M) assumes no shift occurs. The key is to always check Q versus K and set up the ICE table with correct stoichiometric relationships.

This problem involves calculating equilibrium concentrations for H₂(g) + I₂(g) ⇌ 2HI(g). Starting with [H₂] = 0.25 M, [I₂] = 0.25 M, and [HI] = 5.00 M, we calculate Q = [HI]²/([H₂][I₂]) = (5.00)²/(0.25)(0.25) = 400. Since Q > Kc (400 > 100), the reaction must shift left. Setting up the ICE table with [HI] changing by -2x: [H₂] = 0.25 + x, [I₂] = 0.25 + x, and [HI] = 5.00 - 2x at equilibrium. Substituting into Kc = 100: (5.00 - 2x)²/((0.25 + x)²) = 100, taking the square root gives (5.00 - 2x)/(0.25 + x) = 10. Solving: 5.00 - 2x = 2.50 + 10x, which gives x = 0.125, so [HI]eq = 5.00 - 2(0.125) = 4.75 M. A common mistake (choice D, 5.00 M) assumes the initial concentration doesn't change. The key is recognizing that Q > K requires a leftward shift, even with seemingly balanced initial concentrations.

This question asks for equilibrium concentration calculations using an ICE table approach. For H₂(g) + I₂(g) ⇌ 2HI(g) with initial concentrations [H₂] = 0.20 M, [I₂] = 0.20 M, and [HI] = 4.00 M, we first calculate Q = [HI]²/([H₂][I₂]) = (4.00)²/(0.20)(0.20) = 400. Since Q > Kc (400 > 100), the reaction shifts left to reach equilibrium. Setting up the ICE table with [HI] decreasing by 2x gives: [H₂] = 0.20 + x, [I₂] = 0.20 + x, and [HI] = 4.00 - 2x at equilibrium. Substituting into Kc = 100: (4.00 - 2x)²/((0.20 + x)²) = 100, taking the square root yields (4.00 - 2x)/(0.20 + x) = 10. Solving: 4.00 - 2x = 2.00 + 10x, which gives x = 0.10, so [HI]eq = 4.00 - 2(0.10) = 3.80 M. The distractor 4.00 M (choice C) incorrectly assumes no change from initial conditions. The key is recognizing that even small initial reactant concentrations require the system to shift when Q ≠ K.

This question requires calculating equilibrium concentrations for H₂(g) + I₂(g) ⇌ 2HI(g). With initial concentrations [H₂] = 0.40 M, [I₂] = 0.40 M, and [HI] = 2.00 M, we calculate Q = [HI]²/([H₂][I₂]) = (2.00)²/(0.40)(0.40) = 25. Since Q > Kc (25 > 4), the reaction shifts left to reach equilibrium. Using an ICE table where [HI] decreases by 2x: [H₂] = 0.40 + x, [I₂] = 0.40 + x, and [HI] = 2.00 - 2x at equilibrium. Substituting into Kc = 4: (2.00 - 2x)²/((0.40 + x)²) = 4, taking the square root gives (2.00 - 2x)/(0.40 + x) = 2. Solving: 2.00 - 2x = 0.80 + 2x, which yields x = 0.10, so [HI]eq = 2.00 - 2(0.10) = 1.80 M. The distractor 2.00 M (choice C) incorrectly assumes no change occurs. The strategy is to compare Q to K, determine shift direction, and solve using proper stoichiometry.

This problem involves calculating equilibrium concentrations for C(aq) ⇌ D(aq). Setting up the ICE table with [C]₀ = 1.20 M and [D]₀ = 0.010 M, the changes are -x and +x, giving [C] = 1.20 - x and [D] = 0.010 + x at equilibrium. Using Kc = [D]/[C] = 0.030: (0.010 + x)/(1.20 - x) = 0.030. Solving: 0.010 + x = 0.030(1.20 - x) = 0.036 - 0.030x, which gives 1.030x = 0.026, so x = 0.0252. Therefore, [D]eq = 0.010 + 0.0252 = 0.0352 M ≈ 0.036 M. A tempting distractor might be 0.030 M, which incorrectly uses the equilibrium constant as the concentration. The transferable strategy is to always solve for x algebraically and add it to the initial concentration, never confusing Kc with actual concentrations.

This problem requires calculating equilibrium concentrations for H₂(g) + Cl₂(g) ⇌ 2HCl(g). Setting up an ICE table with initial values [H₂]=0.010 M, [Cl₂]=1.00 M, and [HCl]=0 M, we can treat [Cl₂] as constant due to its large initial concentration. The equilibrium expression is Kc = [HCl]²/([H₂][Cl₂]) = 9.0×10⁻⁴. If x mol/L of H₂ reacts, then [HCl]=2x at equilibrium and [H₂]=0.010-x. For small x, the expression becomes (2x)²/(0.010×1.00) = 9.0×10⁻⁴, which gives 4x²/0.010 = 9.0×10⁻⁴, so x² = 2.25×10⁻⁶, and x = 1.5×10⁻³ M. Therefore, [HCl] = 2x = 3.0×10⁻³ M. A common error would be to use [HCl]=x instead of [HCl]=2x in the ICE table, leading to an incorrect answer of 1.5×10⁻³ M. The key strategy is to carefully track stoichiometric coefficients when setting up the ICE table and ensure the equilibrium expression reflects the balanced equation.