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 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. 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.

This question tests the skill of reaction quotient and Le Châtelier's principle. Injecting NO2, a product, increases its concentration, causing the reaction quotient $Q_c = \frac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]}$ 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 NO2 and produces N2O4. This net shift toward reactants restores equilibrium by driving $Q_c$ back to $K_c$. A common misconception is that adding any substance always shifts the equilibrium to the right, but it actually depends on whether the added species is a reactant or product and its impact on Q. Always identify how the stress affects Q compared to K, then predict the shift that drives Q back toward K.

This question tests understanding of reaction quotient and Le Châtelier's principle. When H₂ is added to the equilibrium system, we calculate $Q_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} = \frac{0.30 \times 0.60}{0.30 \times 0.30} = 2.0$, which is greater than $K_c = 1.0$. Since $Q_c > K_c$, the system must shift toward reactants (left) to decrease $Q_c$ back to $K_c$, consuming the excess H₂ that was added. Choice D incorrectly claims that $K_c$ changes when a substance is added—$K_c$ only changes with temperature, not concentration changes. The key strategy is to calculate Q after any stress and compare to K: when $Q > K$, the system shifts left to decrease Q back to K.

This question tests your understanding of reaction quotient and Le Châtelier's principle. When CO is added to the equilibrium system, the concentration of reactants increases, which decreases Q (since [CO] appears in the denominator of the Q expression). With Q = 0.40 and K = 1.0, we have Q < K, meaning there's too little product relative to equilibrium. To reestablish equilibrium, the system must shift to increase Q back to 1.0, which occurs by converting more CO and H₂O into CO₂ and H₂ (shift toward products). A common misconception (choice D) is that adding a reactant doesn't change Q, but adding reactant decreases Q since reactants appear in the denominator. When Q < K, always predict a shift toward products to increase Q back to K.

This question tests your understanding of reaction quotient and Le Châtelier's principle. When HI is removed from the equilibrium system, the concentration of products decreases, which decreases Q (since [HI]² appears in the numerator). With Q = 20 and K = 50, we have Q < K, meaning there's too little product relative to equilibrium. To reestablish equilibrium, the system must shift to increase Q back to 50, which occurs by converting more H₂ and I₂ into HI (shift toward products). A common misconception (choice E) is that removing product makes Q > K, but removing product actually decreases Q since products appear in the numerator. When Q < K, always predict a shift toward products to increase Q back to K.

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 $K$.