This question tests the skill of reaction quotient and Le Châtelier's principle. Removing HI, a product, decreases its concentration, causing the reaction quotient $Q_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}$ to become less than $K_c$. Since $Q_c < K_c$, the system will shift to increase $Q_c$ by favoring the forward reaction, which produces more HI. This net shift toward products restores equilibrium by driving $Q_c$ back to $K_c$. A common misconception is that removing a product causes a shift to the left, but actually, it shifts right to replace the removed species. Always identify how the stress affects $Q$ compared to $K$, then predict the shift that drives $Q$ back toward $K$.

This question tests the skill of reaction quotient and Le Châtelier's principle. Adding CO2, a product, increases its concentration, causing the reaction quotient $Q_c = \frac{[\mathrm{CO_2}][\mathrm{H_2}]}{[\mathrm{CO}][\mathrm{H_2O}]}$ to become greater than $K_c$. Since $Q_c > K_c$, the system will shift to reduce $Q_c$ by favoring the reverse reaction, which consumes CO2 and H2 to produce CO and H2O. This net shift toward reactants restores equilibrium by driving $Q_c$ back to $K_c$. A common misconception is that adding a product always causes no shift, but it actually perturbs $Q$ and triggers a shift to rebalance. Always identify how the stress affects $Q$ compared to $K$, then predict the shift that drives $Q$ back toward $K$.

This question tests the skill of reaction quotient and Le Châtelier's principle. Adding $\mathrm{CH_3COO^-}$, a product, increases its concentration, causing the reaction quotient $Q_c = \frac{[\mathrm{H^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]}$ to become greater than $K_c$. Since $Q_c > K_c$, the system will shift to reduce $Q_c$ by favoring the reverse reaction, which consumes $\mathrm{H^+}$ and $\mathrm{CH_3COO^-}$ to produce $\mathrm{CH_3COOH}$. This net shift toward reactants restores equilibrium by driving $Q_c$ back to $K_c$. A common misconception is that adding a common ion has no effect on weak acid equilibrium, but it actually suppresses dissociation via Le Châtelier's principle. Always identify how the stress affects $Q$ compared to $K$, then predict the shift that drives $Q$ back toward $K$.

This question tests the skill of reaction quotient and Le Châtelier's principle. Removing O2, a reactant, decreases its partial pressure, causing the reaction quotient $Q_p = [SO3]^2 / ([SO2]^2 [O2])$ to become greater than $K_p$ since the denominator decreases. Since $Q_p > K_p$, the system will shift to reduce $Q_p$ by favoring the reverse reaction, which produces more SO2 and O2 while consuming SO3. This net shift toward reactants restores equilibrium by driving $Q_p$ back to $K_p$. A common misconception is that removing a reactant shifts the equilibrium to the right to 'replace' it, but it actually shifts left to produce more of the removed species. Always identify how the stress affects Q compared to K, then predict the shift that drives Q back toward K.

This question tests reaction quotient and Le Châtelier's principle. Removing O2, a reactant, decreases its partial pressure, which makes the denominator smaller in Qp, resulting in Qp = $9.0×10^1$, greater than Kp of $6.0×10^1$. Since Qp > Kp, the system shifts toward the reactants to decrease Qp by forming more NO and O2. This net shift left consumes NO2 to reestablish equilibrium. A common misconception is that removing a reactant makes Qp < Kp and shifts toward products (choice E), but it actually increases Qp, prompting a reverse shift. To predict equilibrium shifts, identify how the stress affects Q, then predict the shift that drives Q back toward K.

This question tests reaction quotient and Le Châtelier's principle. Adding CO, a product, increases its concentration, which raises the numerator in Qc, resulting in Qc = 0.80, greater than Kc of 0.50. Since Qc > Kc, the system shifts toward the reactants to decrease Qc by forming more H2 and CO2. This net shift left consumes H2O and CO to reestablish equilibrium. A common misconception is that adding a product makes Qc < Kc and shifts toward products (choice E), but it actually increases Qc above Kc, prompting a reverse shift. To predict equilibrium shifts, identify how the stress affects Q, then predict the shift that drives Q back toward K.

This question tests understanding of reaction quotient and Le Châtelier's principle. When NO₂ is removed from the equilibrium system, we calculate $Q_c = [\text{NO}_2]^2 / [\text{N}_2\text{O}_4] = (0.30)^2 / (0.40) = 0.225$, which is less than $K_c = 0.50$. Since $Q_c < K_c$, the system must shift toward products (right) to increase $Q_c$ back to $K_c$, producing more NO₂ to replace what was removed. Choice E incorrectly states that $Q_c > K_c$ when a product is removed, which is backwards—removing products always makes $Q < K$. The key strategy is to calculate $Q$ after the stress, compare it to $K$, then predict the shift: if $Q < K$, shift right; if $Q > K$, shift left.

This question tests understanding of reaction quotient and Le Châtelier's principle. After SO₃ is removed, we calculate $Q_p = \frac{(P_{\text{SO}3})^2}{(P{\text{SO}2})^2 \times P{\text{O}_2}} = \frac{(0.40)^2}{(0.50)^2 \times 0.50} = 1.28$, which is less than $K_p = 4.0$. Since $Q_p < K_p$, the system must shift toward products (right) to increase $Q_p$ back to $K_p$, producing more SO₃ to replace what was removed. Choice E incorrectly states that removing product makes $Q_p > K_p$, which is backwards—removing products always decreases Q, making Q < K. The strategy is to calculate Q immediately after the stress: when products are removed, Q decreases below K, so the system shifts right to restore equilibrium.

This question tests understanding of reaction quotient and Le Châtelier's principle. After $\text{FeSCN}^{2+}$ is added, we calculate $Q_c = \frac{[\text{FeSCN}^{2+}]}{[\text{Fe}^{3+}][\text{SCN}^{-}]} = \frac{2.0}{0.10 \times 0.10} = 200$, which is greater than $K_c = 100$. Since $Q_c > K_c$, the system must shift toward reactants (left) to decrease $Q_c$ back to $K_c$, converting some of the added $\text{FeSCN}^{2+}$ back into $\text{Fe}^{3+}$ and $\text{SCN}^{-}$. Choice E incorrectly suggests shifting right because a product was added—the shift direction depends on Q vs K comparison, not simply what was added. The key principle is that adding products increases Q above K, requiring a leftward shift to restore equilibrium.

This question tests understanding of reaction quotient and Le Châtelier's principle. For this heterogeneous equilibrium, solids don't appear in the equilibrium expression, so Q_p = P_CO₂ = 1.6 atm, which is greater than K_p = 0.80 atm. Since Q_p > K_p, the system must shift toward reactants (left) to decrease the CO₂ pressure back to equilibrium, converting some CO₂ back into CaCO₃. Choice D incorrectly claims that solids make Q_p constant—solids are omitted from Q and K expressions, but Q still varies with gas pressures. The key insight is that for heterogeneous equilibria, only gases and aqueous species appear in Q and K expressions.