Direction of Reversible Reactions - College Chemistry

1. For the equilibrium $2NO_2(g) ⇌ N_2O_4(g)$, $K_p = 6.8$ at 45°C. A reaction mixture has partial pressures $P_{NO_2} = 0.35$ atm and $P_{N_2O_4} = 0.82$ atm. To predict the direction of reaction, a student calculates $Q_p = \frac{0.82}{(0.35)^2} = 6.7$. What conclusion should the student draw?
   Answer: Since Q ≈ K, the system is essentially at equilibrium with minimal net reaction When you encounter equilibrium problems involving reaction quotients, you're comparing the current state of a system to its equilibrium state to predict which direction the reaction will proceed. The student correctly calculated the reaction quotient: $Q_p = \frac{P_{N_2O_4}}{P_{NO_2}^2} = \frac{0.82}{(0.35)^2} = 6.7$. Since $K_p = 6.8$ at this temperature, we have $Q_p = 6.7$ and $K_p = 6.8$. These values are essentially equal (within experimental uncertainty), indicating the system is very close to equilibrium with minimal driving force in either direction. Answer A correctly identifies this situation. Answer B is wrong because $Q < K$ would normally drive the forward reaction, but the difference here (6.7 vs 6.8) is negligible—certainly not "strongly" in either direction. Answer C incorrectly states that $Q > K$, which isn't true since 6.7 < 6.8. Even if it were true, the tiny difference wouldn't cause a "strong" reverse reaction. Answer D is incorrect because temperature doesn't appear in the $Q_p$ calculation itself—temperature affects the value of $K_p$, but once you know $K_p$ at a given temperature, you can compare it directly to $Q_p$. Remember that when $Q_p$ and $K_p$ are very close in value (typically within 5-10% of each other), treat the system as essentially at equilibrium rather than predicting significant reaction in either direction. Focus on the magnitude of the difference, not just which is larger.
2. For the reaction $N_2O_4(g) ⇌ 2NO_2(g)$, $K_c = 4.63 × 10^{-3}$ at 25°C. A reaction mixture contains 0.0236 M $N_2O_4$ and 0.0318 M $NO_2$. What is the direction of the net reaction?
   Answer: The reaction will proceed in the reverse direction When you encounter equilibrium problems asking about reaction direction, you need to compare the reaction quotient (Q) with the equilibrium constant (K). This comparison tells you which way the reaction must shift to reach equilibrium. First, calculate the reaction quotient using the same expression as the equilibrium constant, but with current concentrations instead of equilibrium concentrations: $Q_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(0.0318)^2}{(0.0236)} = \frac{0.00101}{0.0236} = 0.0428$ Now compare Q to the given $K_c = 4.63 × 10^{-3} = 0.00463$: Since $Q_c = 0.0428 > K_c = 0.00463$, the reaction quotient is greater than the equilibrium constant. This means there's too much product ($NO_2$) relative to reactant ($N_2O_4$) compared to the equilibrium ratio. The reaction must shift left (reverse direction) to decrease the numerator and increase the denominator until Q equals K. Choice A is correct because Q > K indicates reverse reaction direction. Choice B is wrong because Q > K never indicates forward direction (that only happens when Q < K). Choice C is incorrect because the reaction is at equilibrium only when Q = K, which isn't the case here. Choice D is wrong because comparing Q and K always determines reaction direction when both values and concentrations are known. Study tip: Remember the Q vs K rule: Q < K means forward, Q > K means reverse, Q = K means equilibrium. Always calculate Q first, then compare to make your prediction.
3. The water-gas shift reaction $CO(g) + H_2O(g) ⇌ CO_2(g) + H_2(g)$ reaches equilibrium with $K_c = 0.64$ at 800°C. If a reactor initially contains 2.0 M CO, 2.0 M $H_2O$, 1.0 M $CO_2$, and 1.0 M $H_2$, what happens immediately after startup?
   Answer: The reaction proceeds forward because Q = 0.25 < K = 0.64 When you encounter equilibrium problems with given concentrations, you need to determine the reaction direction by comparing the reaction quotient (Q) to the equilibrium constant (K). First, calculate Q using the same expression as the equilibrium constant: $Q_c = \frac{[CO_2][H_2]}{[CO][H_2O]}$. Substituting the initial concentrations: $Q_c = \frac{(1.0)(1.0)}{(2.0)(2.0)} = \frac{1.0}{4.0} = 0.25$. Since Q = 0.25 and K = 0.64, we have Q < K. When Q < K, the reaction must proceed forward to reach equilibrium, converting more reactants into products until Q equals K. Looking at the answer choices: Choice A correctly identifies that Q = 0.25 < K = 0.64, leading to a forward reaction. Choice B incorrectly calculates Q as 4.0 – this happens when you flip the fraction and put reactants in the numerator instead of products. Choice C wrongly assumes equal concentrations mean equilibrium, but equilibrium depends on the specific ratio defined by K, not equal amounts. Choice D correctly calculates Q = 0.25 < K = 0.64 but incorrectly concludes the reaction goes in reverse – when Q < K, the reaction always proceeds forward. Remember this pattern: Q < K means forward reaction, Q > K means reverse reaction, and Q = K means equilibrium. Always double-check your Q expression matches the K expression with products over reactants.
4. A student analyzes the equilibrium $SO_2(g) + \frac{1}{2}O_2(g) ⇌ SO_3(g)$ with $K_p = 128$ at 627°C. The partial pressures are $P_{SO_2} = 0.40$ atm, $P_{O_2} = 0.60$ atm, and $P_{SO_3} = 25$ atm. The student calculates $Q_p = \frac{25}{(0.40)(0.60)^{0.5}} = \frac{25}{0.310} = 80.6$. What should the student conclude about the reaction direction?
   Answer: The reaction will proceed forward because Q < K When you encounter equilibrium problems involving reaction quotient (Q) and equilibrium constant (K), you're determining which direction a reaction will proceed to reach equilibrium. The key principle is comparing Q to K. The student correctly calculated $Q_p = 80.6$ using the equilibrium expression $Q_p = \frac{P_{SO_3}}{P_{SO_2} \cdot P_{O_2}^{0.5}}$. Since $Q_p = 80.6$ and $K_p = 128$, we have Q < K. When Q < K, the reaction quotient is smaller than it should be at equilibrium, meaning there's too little product relative to reactants. The reaction must shift forward (toward products) to increase Q until it equals K. Looking at the wrong answers: Choice B incorrectly states that Q > K, which is mathematically false since 80.6 < 128. Choice C makes the right comparison (Q < K) but draws the wrong conclusion about reaction direction—when Q < K, the reaction proceeds forward, not reverse. Choice D suggests the system is near equilibrium, but with Q and K differing by nearly 50 units (128 - 80.6 = 47.4), this represents a significant deviation from equilibrium. Choice A correctly identifies both the relationship (Q < K) and the proper conclusion (forward reaction). Study tip: Remember the Q vs K rule: Q < K means "not enough product" so the reaction goes forward; Q > K means "too much product" so the reaction goes reverse. The reaction always shifts to make Q equal to K.
5. Consider the equilibrium $COCl_2(g) ⇌ CO(g) + Cl_2(g)$ with $K_p = 8.3 × 10^{-15}$ at 25°C. A container initially holds 0.50 atm $COCl_2$, 0.10 atm CO, and 0.10 atm $Cl_2$. What is the direction of the spontaneous reaction?
   Answer: Reverse, because Q = 2.0 × $10^{-2}$ > K When you encounter an equilibrium problem asking about reaction direction, you need to compare the reaction quotient Q to the equilibrium constant K. The reaction will proceed in whichever direction brings Q closer to K. First, calculate the reaction quotient using the same expression as the equilibrium constant, but with initial pressures instead of equilibrium pressures: $Q_p = \frac{P_{CO} \times P_{Cl_2}}{P_{COCl_2}} = \frac{(0.10)(0.10)}{0.50} = \frac{0.010}{0.50} = 2.0 × 10^{-2}$ Now compare Q to K: $Q_p = 2.0 × 10^{-2} = 0.02$ while $K_p = 8.3 × 10^{-15}$. Since Q is much larger than K, the reaction must shift in the reverse direction to decrease Q until it equals K. Choice B correctly identifies both the calculated Q value and the reverse direction. Choice A has the right Q calculation but wrong direction—when Q > K, the reaction goes reverse, not forward. Choice C uses an incorrect Q calculation ($5.0 × 10^{-2}$), likely from a math error in the division. Choice D has the same incorrect Q value as C, though it does correctly identify the reverse direction. Remember this pattern: Q > K means reverse reaction, Q < K means forward reaction, Q = K means equilibrium. Also, always double-check your Q calculation—make sure products are in the numerator and reactants in the denominator, each raised to their stoichiometric coefficients.
6. The equilibrium $C(s) + CO_2(g) ⇌ 2CO(g)$ has $K_p = 1.52$ at 700°C. A reaction vessel contains solid carbon with $P_{CO_2} = 0.50$ atm and $P_{CO} = 1.10$ atm. What change will occur as the system approaches equilibrium?
   Answer: $P_{CO}$ will decrease and $P_{CO_2}$ will increase as the reaction proceeds reverse When you encounter equilibrium problems, you need to compare the reaction quotient (Q) with the equilibrium constant (K) to predict which direction the reaction will proceed. First, calculate the reaction quotient using the given pressures. For this reaction, $Q_p = \frac{P_{CO}^2}{P_{CO_2}}$ (note that solid carbon doesn't appear in the expression). Substituting the values: $Q_p = \frac{(1.10)^2}{0.50} = \frac{1.21}{0.50} = 2.42$ Since $Q_p = 2.42 > K_p = 1.52$, the system has too much product relative to reactant compared to equilibrium. The reaction must shift left (reverse direction) to reach equilibrium, consuming CO and producing $CO_2$. Answer B correctly identifies this: $P_{CO}$ will decrease and $P_{CO_2}$ will increase as the reaction proceeds in reverse. Answer A describes the forward reaction, which would occur if $Q_p < K_p$, but that's not the case here. Answer C incorrectly suggests that solid carbon "buffers" the system—while solids don't affect equilibrium expressions, they don't prevent pressure changes in the gas phase. Answer D is chemically impossible since both gas pressures cannot increase simultaneously in this closed system without violating mass balance. Study tip: Always calculate Q first and compare it to K. If Q > K, the reaction shifts left (reverse); if Q < K, it shifts right (forward). Remember that solids and pure liquids don't appear in equilibrium expressions, but they don't prevent the reaction from proceeding.
7. For the equilibrium $2NO_2(g) ⇌ N_2O_4(g)$, $K_p = 6.7$ at 45°C. A reaction vessel initially contains only $NO_2$ at 0.50 atm partial pressure. After some time, $P_{NO_2} = 0.32$ atm and $P_{N_2O_4} = 0.09$ atm. What is occurring in this system at this moment?
   Answer: The reaction is proceeding toward equilibrium in the forward direction When you encounter equilibrium problems asking about the direction of reaction, you need to compare the reaction quotient (Q) to the equilibrium constant (K) to determine which way the reaction will proceed. First, calculate the reaction quotient using the given partial pressures. For $2NO_2(g) ⇌ N_2O_4(g)$, the expression is $Q_p = \frac{P_{N_2O_4}}{(P_{NO_2})^2}$. Substituting the values: $Q_p = \frac{0.09}{(0.32)^2} = \frac{0.09}{0.1024} = 0.88$ Now compare Q to the given $K_p = 6.7$. Since $Q_p = 0.88 < K_p = 6.7$, the reaction must shift forward to reach equilibrium, producing more $N_2O_4$ and consuming more $NO_2$. Looking at the wrong answers: Answer B suggests the reverse direction, but since Q < K, the forward reaction is favored. Answer C claims equilibrium has been reached, but this would only be true if Q equaled K (6.7), not 0.88. Answer D incorrectly assumes the reaction stops due to insufficient reactant, but equilibrium reactions continue in both directions even when reactant concentrations are low. The correct answer is A because when Q < K, the reaction proceeds forward until Q increases to equal K. Study tip: Remember the Q vs K rule: Q < K means go forward, Q > K means go reverse, Q = K means equilibrium. Always calculate Q first, then compare to the given K value to predict reaction direction.
8. The decomposition reaction $2HI(g) ⇌ H_2(g) + I_2(g)$ has $K_c = 1.26 × 10^{-3}$ at 500°C. A sealed flask initially contains only 0.500 M HI. After some time, analysis shows [HI] = 0.482 M, [$H_2$] = 0.009 M, and [$I_2$] = 0.009 M. What is happening in this system?
   Answer: The reaction is proceeding toward equilibrium in the forward direction When you encounter equilibrium problems involving concentration changes, you need to determine whether the system has reached equilibrium by comparing the reaction quotient (Q) to the equilibrium constant (K). First, calculate the reaction quotient using the given concentrations: $Q_c = \frac{[H_2][I_2]}{[HI]^2} = \frac{(0.009)(0.009)}{(0.482)^2} = \frac{8.1 × 10^{-5}}{0.232} = 3.49 × 10^{-4}$ Since $Q_c = 3.49 × 10^{-4}$ is less than $K_c = 1.26 × 10^{-3}$, the reaction must continue in the forward direction to reach equilibrium. When Q < K, the system shifts right to increase the numerator (products) and decrease the denominator (reactants) until Q equals K. Answer A is correct because Q < K indicates the forward reaction will continue. Answer B is wrong because the reverse direction would occur if Q > K, which isn't the case here. Answer C is incorrect because Q ≠ K, so equilibrium hasn't been reached. Answer D reflects a common misconception—reactions don't "stop" due to insufficient reactants unless one is completely consumed. The system will continue shifting until equilibrium is established, regardless of how much starting material remains. Remember this key relationship: when Q < K, the reaction proceeds forward; when Q > K, it proceeds in reverse; when Q = K, you're at equilibrium. Always calculate Q first when analyzing whether an equilibrium system is at rest or still shifting.
9. A student is investigating the equilibrium $H_2S(g) + I_2(s) ⇌ 2HI(g) + S(s)$ at 60°C, where $K_c = 1.34 × 10^{-5}$. The concentrations are [$H_2S$] = 0.045 M and [HI] = 0.0012 M, with solid $I_2$ and S present. What will happen to the [HI] as the system approaches equilibrium?
   Answer: [HI] will decrease because Q = 3.2 × $10^{-5}$ > K When you encounter equilibrium problems asking about concentration changes, you need to compare the reaction quotient (Q) to the equilibrium constant (K) to predict the direction of change. First, calculate Q using the same expression as K, but with current concentrations. For this reaction, $Q_c = \frac{[HI]^2}{[H_2S]}$ (solids are excluded from equilibrium expressions). Substituting the given values: $Q_c = \frac{(0.0012)^2}{0.045} = \frac{1.44 × 10^{-6}}{0.045} = 3.2 × 10^{-5}$ Since $Q_c = 3.2 × 10^{-5}$ is greater than $K_c = 1.34 × 10^{-5}$, the reaction must shift left (toward reactants) to reach equilibrium. This means [HI] will decrease as the system converts HI back to reactants. Option A incorrectly states that [HI] will increase despite correctly calculating Q > K. When Q > K, the reaction always shifts left, decreasing product concentrations. Option C makes the opposite error, correctly predicting an increase in [HI] but falsely claiming Q < K. Option D uses an incorrect Q calculation ($8.0 × 10^{-6}$) that's smaller than K, leading to the wrong directional prediction. Remember this key relationship: when Q > K, the reaction shifts left (products decrease); when Q < K, the reaction shifts right (products increase). Always calculate Q carefully using the correct equilibrium expression, excluding any solids or pure liquids from your calculation.
10. The equilibrium $2IBr(g) ⇌ I_2(g) + Br_2(g)$ has $K_c = 8.5 × 10^{-3}$ at 150°C. In a sealed flask, [IBr] = 0.25 M, [$I_2$] = 0.10 M, and [$Br_2$] = 0.080 M. Which statement correctly describes the immediate behavior of this system?
    Answer: The system will shift toward reactants because Q = 1.28 × $10^{-2}$ > K When you encounter an equilibrium problem asking about system behavior, you need to compare the reaction quotient (Q) to the equilibrium constant (K) to predict which direction the reaction will proceed. First, calculate the reaction quotient Q using the same expression as the equilibrium constant, but with the current concentrations: $Q_c = \frac{[I_2][Br_2]}{[IBr]^2} = \frac{(0.10)(0.080)}{(0.25)^2} = \frac{0.008}{0.0625} = 1.28 × 10^{-2}$ Since $Q_c = 1.28 × 10^{-2}$ and $K_c = 8.5 × 10^{-3}$, we have Q > K. When Q > K, the system has too much product relative to equilibrium, so it shifts toward reactants to decrease Q until it equals K. Looking at the answer choices: Choice A correctly identifies both the Q value (1.28 × $10^{-2}$) and the proper conclusion that the system shifts toward reactants because Q > K. Choice B uses the correct Q value but incorrectly states Q < K, leading to the wrong directional prediction. Choice C calculates Q incorrectly as 3.2 × $10^{-2}$ (likely from arithmetic errors) but reaches the correct directional conclusion by chance. Choice D combines the incorrect Q calculation with an impossible conclusion—it states Q > K but predicts a shift toward products, which contradicts equilibrium principles. Remember: Q > K means shift left (toward reactants), Q < K means shift right (toward products). Always double-check your Q calculation since small arithmetic errors can lead to wrong predictions about equilibrium direction.
11. Consider the equilibrium $Fe^{3+}(aq) + SCN^-(aq) ⇌ FeSCN^{2+}(aq)$ with $K_c = 138$ at room temperature. A solution is prepared by mixing equal volumes of 0.002 M $Fe^{3+}$ and 0.002 M $SCN^-$. Immediately after mixing, what is the direction of the net reaction?
    Answer: Forward, because no product is initially present so Q = 0 < K When you encounter equilibrium problems asking about reaction direction, you need to compare the reaction quotient Q to the equilibrium constant K. The reaction will proceed in whichever direction brings Q closer to K. To find Q, you use the same expression as K but with initial concentrations. Here, $Q_c = \frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^-]}$. Since equal volumes are mixed, each concentration is halved: both $[Fe^{3+}]$ and $[SCN^-]$ become 0.001 M, while $[FeSCN^{2+}] = 0$ initially. Therefore: $Q_c = \frac{0}{(0.001)(0.001)} = 0$ Since $Q_c = 0 < K_c = 138$, the reaction must proceed forward to increase Q until it equals K. Answer A correctly identifies that Q = 0 because no product is present initially, and since Q < K, the forward reaction occurs. Answer B incorrectly suggests the reaction goes reverse - but when Q < K, reactions always go forward, regardless of concentration magnitude. Answer C wrongly assumes stoichiometric proportions prevent net reaction - but equilibrium position depends on K, not initial ratios. Answer D makes an error in calculating Q, claiming Q = 1 due to equal concentrations, which ignores that the product concentration is zero. Remember: always calculate Q first using initial concentrations, then compare to K. If Q < K, the reaction goes forward; if Q > K, it goes reverse. The key insight is that Q = 0 whenever products are initially absent.
12. At 1000 K, the equilibrium constant for $CO(g) + H_2O(g) ⇌ CO_2(g) + H_2(g)$ is $K_c = 1.56$. If a reaction vessel contains 0.100 M CO, 0.100 M $H_2O$, 0.200 M $CO_2$, and 0.200 M $H_2$, which statement correctly describes the system?
    Answer: Q = 4.00, so the reaction will proceed to the left When you encounter equilibrium problems with given concentrations, you need to compare the reaction quotient (Q) to the equilibrium constant (K) to determine which direction the reaction will proceed. First, calculate Q using the same expression as the equilibrium constant: $Q_c = \frac{[CO_2][H_2]}{[CO][H_2O]}$. Substituting the given concentrations: $Q_c = \frac{(0.200)(0.200)}{(0.100)(0.100)} = \frac{0.0400}{0.0100} = 4.00$. Since $Q_c = 4.00$ is greater than $K_c = 1.56$, the system has too much product relative to reactants compared to equilibrium. To reach equilibrium, the reaction must shift left (toward reactants) to decrease the numerator and increase the denominator until Q equals K. Looking at the answer choices: A correctly identifies Q = 4.00 and the leftward shift. B incorrectly calculates Q as 0.25 (likely from inverting the expression) and wrongly predicts a rightward shift. C correctly notes that when Q = K the system is at equilibrium, but Q ≠ 1.56 here. D correctly calculates Q = 4.00 but incorrectly concludes the reaction proceeds right—when Q > K, the reaction always shifts left to reduce the product-to-reactant ratio. Remember this key relationship: when Q > K, shift left; when Q < K, shift right; when Q = K, you're at equilibrium. Always double-check your Q calculation by ensuring products are in the numerator and reactants in the denominator, just like the K expression.
13. A student is studying the equilibrium $H_2(g) + I_2(g) ⇌ 2HI(g)$ with $K_c = 54.3$ at 700 K. The student prepares a mixture with [$H_2$] = 0.100 M, [$I_2$] = 0.200 M, and [HI] = 1.50 M, then predicts the direction using $Q_c = \frac{(1.50)^2}{(0.100)(0.200)} = 112.5$. What should the student conclude?
    Answer: Since Q > K, the reverse reaction is favored and [HI] will decrease When you encounter equilibrium problems involving reaction quotients, you're being tested on your ability to predict which direction a reaction will proceed to reach equilibrium. The student correctly calculated the reaction quotient: $Q_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(1.50)^2}{(0.100)(0.200)} = 112.5$. Since $Q_c = 112.5 > K_c = 54.3$, the system has too much product (HI) relative to reactants compared to the equilibrium ratio. To reach equilibrium, the reaction must shift left (reverse direction) to consume HI and produce more $H_2$ and $I_2$. This means [HI] will decrease. Looking at the wrong answers: Choice B incorrectly states that when Q > K, the forward reaction is favored. This is backwards—when Q > K, there's excess product, so the reverse reaction must be favored. Choice C correctly identifies that the reverse reaction is favored and [HI] will decrease, but wrongly claims Q < K. The calculation clearly shows Q > K. Choice D suggests a calculation error involving temperature, but temperature is already incorporated into the given $K_c$ value of 54.3 at 700 K, and the student's math is correct. Remember this key relationship: when Q > K, the reaction shifts left (reverse); when Q < K, it shifts right (forward). Always compare your calculated Q value to the given K value first, then determine the direction of shift based on whether you have excess products or reactants.
14. For the gas-phase reaction $A_2(g) + 3B_2(g) ⇌ 2AB_3(g)$, $K_p = 4.5 × 10^4$ at 400 K. A reaction vessel contains $P_{A_2} = 0.10$ atm, $P_{B_2} = 0.30$ atm, and $P_{AB_3} = 15$ atm. A student claims the reaction will proceed in the reverse direction. Is this claim correct?
    Answer: Yes, because Q = 8.3 × $10^4$ > K, so the reaction proceeds reverse When you encounter equilibrium problems asking about reaction direction, you need to compare the reaction quotient (Q) to the equilibrium constant (K). This tells you which way the reaction will shift to reach equilibrium. First, calculate Q using the same expression as K, but with the given pressures instead of equilibrium pressures: $Q_p = \frac{(P_{AB_3})^2}{(P_{A_2})(P_{B_2})^3} = \frac{(15)^2}{(0.10)(0.30)^3} = \frac{225}{(0.10)(0.027)} = \frac{225}{0.0027} = 8.3 × 10^4$ Since $Q_p = 8.3 × 10^4 > K_p = 4.5 × 10^4$, the reaction quotient exceeds the equilibrium constant. This means there are too many products relative to reactants compared to the equilibrium position, so the reaction must proceed in the reverse direction to reach equilibrium. Answer A incorrectly calculates Q as $1.85 × 10^4$ (likely from computational error) and draws the wrong conclusion about reaction direction. Answer C ignores the quantitative relationship between Q and K, making a qualitative guess based on "high product pressure" without proper calculation. Answer D gets the wrong Q value (same error as A) but would have the correct reasoning if the calculation were right. Remember: when Q > K, the reaction shifts left (reverse); when Q < K, it shifts right (forward); when Q = K, it's at equilibrium. Always calculate Q carefully using the correct stoichiometric coefficients as exponents.
15. For the equilibrium $3A(g) + B(g) ⇌ 2C(g) + D(g)$, $K_c = 0.85$ at 450 K. A reaction mixture has [A] = 1.2 M, [B] = 0.80 M, [C] = 0.60 M, and [D] = 0.40 M. A student needs to predict whether the concentrations of products will increase or decrease. What is the correct prediction?
    Answer: Product concentrations will increase because Q = 0.104 < K When you encounter equilibrium problems asking about concentration changes, you need to compare the reaction quotient (Q) to the equilibrium constant (K) to determine which direction the reaction will proceed. First, calculate Q using the same expression as K, but with current concentrations instead of equilibrium concentrations. For this reaction, $Q = \frac{[C]^2[D]}{[A]^3[B]}$. Substituting the given values: $Q = \frac{(0.60)^2(0.40)}{(1.2)^3(0.80)} = \frac{0.144}{1.382} = 0.104$ Since Q = 0.104 and K = 0.85, we have Q < K. This means the reaction must shift right (toward products) to reach equilibrium, so product concentrations will increase. Looking at the wrong answers: Option B incorrectly states that Q > K when Q is actually less than K. Options C and D both calculate Q incorrectly as 1.44 - this likely comes from flipping the Q expression upside down, putting reactants in the numerator and products in the denominator. Even if Q were 1.44, option C would be wrong because when Q > K, the reaction shifts left (decreasing products), not right. The correct answer is A: product concentrations will increase because Q = 0.104 < K. Remember this key relationship: when Q < K, the reaction shifts right toward products; when Q > K, it shifts left toward reactants. Always double-check your Q expression matches the equilibrium constant format with products over reactants, each raised to their stoichiometric coefficients.
16. The equilibrium $CaCO_3(s) ⇌ CaO(s) + CO_2(g)$ has $K_p = 1.16$ atm at 900°C. A sealed container holds solid $CaCO_3$ and solid CaO with $P_{CO_2} = 2.3$ atm. What will occur when the system reaches equilibrium?
    Answer: $CO_2$ pressure will decrease as $CaCO_3$ forms from CaO When you encounter equilibrium problems involving both solids and gases, focus on comparing the reaction quotient (Q) to the equilibrium constant (K) to predict which direction the reaction will proceed. Here, the equilibrium constant $K_p = 1.16$ atm represents the CO₂ pressure when the system is at equilibrium. Since only gases appear in the equilibrium expression for heterogeneous equilibria, $K_p = P_{CO_2}$. Currently, $P_{CO_2} = 2.3$ atm, which is greater than $K_p = 1.16$ atm. When the actual pressure exceeds the equilibrium pressure, the system must shift left (toward reactants) to reduce the CO₂ pressure. This means CO₂ will react with CaO to form more CaCO₃, decreasing the CO₂ pressure until it reaches 1.16 atm. Looking at the wrong answers: Choice B suggests the opposite direction, but since $P_{CO_2} > K_p$, decomposition would move the system further from equilibrium. Choice C incorrectly assumes solids prevent equilibrium changes—while solids don't appear in the equilibrium expression, the reaction can still proceed in either direction. Choice D misunderstands equilibrium; having all species present doesn't guarantee equilibrium—you must compare Q to K. Remember this pattern: when $P_{CO_2} > K_p$, the reverse reaction occurs to reduce gas pressure; when $P_{CO_2} < K_p$, the forward reaction increases gas pressure. The equilibrium constant tells you exactly what the final gas pressure must be.
17. For the equilibrium $2SO_3(g) ⇌ 2SO_2(g) + O_2(g)$, $K_p = 1.8 × 10^{-5}$ at 600°C. A mixture has $P_{SO_3} = 2.0$ atm, $P_{SO_2} = 0.050$ atm, and $P_{O_2} = 0.025$ atm. What is the direction of the net reaction?
    Answer: Forward, because Q = 1.56 × $10^{-5}$ < K When you encounter equilibrium problems asking about reaction direction, you need to compare the reaction quotient (Q) to the equilibrium constant (K). This tells you whether the system is at equilibrium or which way it will shift to reach equilibrium. First, calculate Q using the same expression as K, but with current pressures instead of equilibrium pressures. For this reaction, $Q_p = \frac{P_{SO_2}^2 \times P_{O_2}}{P_{SO_3}^2}$ Substituting the given values: $Q_p = \frac{(0.050)^2 \times (0.025)}{(2.0)^2} = \frac{0.0025 \times 0.025}{4.0} = \frac{6.25 \times 10^{-5}}{4.0} = 1.56 \times 10^{-5}$ Since $Q_p = 1.56 \times 10^{-5} < K_p = 1.8 \times 10^{-5}$, the reaction will proceed forward to increase Q until it equals K. Looking at the wrong answers: Answer B incorrectly states that $1.56 \times 10^{-5} > K$, which is mathematically wrong. Answer C gets the wrong Q value ($6.4 \times 10^{-6}$), likely from calculation errors in the numerator or denominator. Answer D combines both mistakes—the wrong Q value and incorrect comparison. Study tip: Remember the Q vs. K rules: when Q < K, the reaction goes forward; when Q > K, it goes reverse; when Q = K, it's at equilibrium. Double-check your Q calculation by ensuring you're using the correct stoichiometric coefficients as exponents and placing products over reactants.
18. A researcher examines the equilibrium $CH_4(g) + H_2O(g) ⇌ CO(g) + 3H_2(g)$ at 1000 K, where $K_p = 3.2$. The partial pressures are $P_{CH_4} = 0.80$ atm, $P_{H_2O} = 1.2$ atm, $P_{CO} = 0.45$ atm, and $P_{H_2} = 1.5$ atm. Based on these conditions, what will happen to the partial pressure of $H_2$?
    Answer: $P_{H_2}$ will increase because Q = 1.6 < K When you encounter equilibrium problems asking about the direction of reaction, you need to compare the reaction quotient Q to the equilibrium constant K to predict which way the system will shift. First, calculate Q using the same expression as K, but with current partial pressures instead of equilibrium values. For this reaction, $Q_p = \frac{P_{CO} \times P_{H_2}^3}{P_{CH_4} \times P_{H_2O}}$. Substituting the given values: $Q_p = \frac{(0.45)(1.5)^3}{(0.80)(1.2)} = \frac{(0.45)(3.375)}{0.96} = 1.58 ≈ 1.6$. Since Q = 1.6 < K = 3.2, the reaction must shift forward (to the right) to reach equilibrium. This means more products will form, including more $H_2$, so the partial pressure of $H_2$ will increase. This confirms answer A is correct. Looking at the wrong answers: B incorrectly states that Q > K when we calculated Q < K. C and D both claim Q = 3.5, which represents a calculation error—likely from miscalculating $P_{H_2}^3 = (1.5)^3 = 3.375$ somewhere in the arithmetic. Additionally, D combines this wrong Q value with the wrong prediction about pressure change direction. Remember this key relationship: when Q < K, the reaction shifts forward (toward products), and when Q > K, it shifts backward (toward reactants). Always double-check your Q calculation, especially when dealing with coefficients greater than 1—they become exponents in the expression.

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