This question tests understanding of how equilibrium constants change when a reaction is reversed. The original reaction H₂(g) + I₂(g) ⇌ 2HI(g) has equilibrium constant K = [HI]²/([H₂][I₂]). The new reaction 2HI(g) ⇌ H₂(g) + I₂(g) is the exact reverse of the original. When a reaction is reversed, the new equilibrium constant is always the reciprocal of the original: K_new = 1/K. Students might incorrectly choose 1/K² (choice E), thinking the coefficient 2 in front of HI affects the relationship. Remember: reversing a reaction always gives K_new = 1/K, regardless of any coefficients in the balanced equation.

This question assesses halving a reaction's equilibrium constant. The original is 2NO(g) + Cl₂(g) ⇌ 2NOCl(g) with K = [NOCl]² / ([NO]²[Cl₂]), and the new is halved, NO(g) + ½Cl₂(g) ⇌ NOCl(g). K_new = $K^{1/2}$ = √K, as the expression takes the square root. This follows scaling by 1/2. A tempting distractor is D, K², incorrect for using the full power instead of half, stemming from inverting the factor. Calculate the scaling n and use K_new = $K^n$ for same-direction reactions.

This question assesses the equilibrium constant for a reversed reaction. The original is CO₂(g) + H₂(g) ⇌ CO(g) + H₂O(g) with K = [CO][H₂O] / ([CO₂][H₂]), and the new is the reverse. Thus, K_new = 1/K, as the expression inverts. This holds for balanced reactions with equal terms. A tempting distractor is D, 1/K², incorrect if scaling is mistakenly added, stemming from overcomplicating simple reversal. Identify pure reversal by swapped sides and directly use K_new = 1/K.

This question tests understanding of how equilibrium constants change when a reaction is both reversed and coefficients are multiplied. The original reaction CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) has equilibrium constant K. The new reaction 2CO₂(g) + 2H₂(g) ⇌ 2CO(g) + 2H₂O(g) involves two changes: reversing the reaction (which gives 1/K) and doubling all coefficients (which squares the result). Therefore, K_new = (1/K)² = 1/K². Students often incorrectly choose 1/K (choice A), forgetting to account for the coefficient change. Remember: when both reversing and changing coefficients, apply both transformations: reverse first, then adjust for coefficient changes.

This question tests understanding of how equilibrium constants change when a reaction is reversed. The original reaction 2NO₂(g) ⇌ N₂O₄(g) has equilibrium constant K = [N₂O₄]/[NO₂]². The new reaction N₂O₄(g) ⇌ 2NO₂(g) is the exact reverse of the original. When a reaction is reversed, the new equilibrium constant is always the reciprocal: K_new = [NO₂]²/[N₂O₄] = 1/K. Students might incorrectly choose 1/K² (choice A), thinking the coefficient 2 affects the reciprocal relationship. Remember: reversing a reaction always gives K_new = 1/K, regardless of coefficients in the balanced equation.

This question tests reversal in a heterogeneous system. The original is C(s) + CO₂(g) ⇌ 2CO(g) with K = [CO]² / [CO₂], and the new is the reverse, 2CO(g) ⇌ C(s) + CO₂(g). K_new = 1/K, inverting the expression. Solids are omitted, preserving the relationship. A tempting distractor is D, 1/K², wrong for extra powering, due to misapplying the coefficient of CO. Confirm exact reversal and use K_new = 1/K, checking coefficients match.

This question tests understanding of how equilibrium constants change when all coefficients are multiplied by the same factor. The original reaction PCl₅ ⇌ PCl₃ + Cl₂ has K = [PCl₃][Cl₂]/[PCl₅]. When all coefficients are doubled, K_new = [PCl₃]²[Cl₂]²/[PCl₅]² = ([PCl₃][Cl₂]/[PCl₅])² = K². Therefore, K_new = K², making choice C correct. A common misconception is thinking that doubling coefficients has no effect on K (choosing K_new = K), but each concentration term is raised to the power of its coefficient. When coefficients are multiplied by n, raise the original K to the nth power.

This question tests understanding of how equilibrium constants change when all coefficients in a reaction are multiplied by the same factor. The original reaction H₂ + I₂ ⇌ 2HI has K = [HI]²/([H₂][I₂]). When all coefficients are doubled, the new equilibrium expression becomes K_new = [HI]⁴/([H₂]²[I₂]²) = ([HI]²/([H₂][I₂]))² = K². Therefore, K_new = K², making choice D correct. Students might incorrectly think doubling coefficients doubles K (which would give K_new = 2K, not an option), but the relationship is exponential, not linear. When all coefficients are multiplied by n, the new equilibrium constant equals $K^n$.

This question tests understanding of equilibrium constants for combined reactions. When adding A(g) + B(g) ⇌ C(g) with K₁ = K to its reverse C(g) ⇌ A(g) + B(g) with K₂ = 1/K, the net equation is 0 ⇌ 0. For combined reactions, equilibrium constants multiply: K_new = K₁ × K₂ = K × (1/K) = 1. This makes physical sense because the net equation represents no net change, and at equilibrium, there's no driving force in either direction. Students who choose option B (1/K) might think only one reaction's K matters. The key principle is that when reactions are added, their equilibrium constants multiply, and K × (1/K) always equals 1.

This question tests understanding of how equilibrium constants change when reaction coefficients are multiplied. The original reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) has equilibrium constant K = [PCl₃][Cl₂]/[PCl₅]. When all coefficients are doubled to get 2PCl₅(g) ⇌ 2PCl₃(g) + 2Cl₂(g), the new equilibrium expression becomes K_new = [PCl₃]²[Cl₂]²/[PCl₅]² = ([PCl₃][Cl₂]/[PCl₅])² = K². This follows the rule that when coefficients are multiplied by n, the equilibrium constant is raised to the nth power. A common error is thinking the equilibrium constant remains unchanged (choice E), failing to recognize that the exponents in the equilibrium expression change with the coefficients. To determine the new K value, identify the coefficient multiplier and raise the original K to that power.