Calculating the Equilibrium Constant - College Chemistry

1. At equilibrium for the reaction $4NH_3(g) + 5O_2(g) \rightleftharpoons 4NO(g) + 6H_2O(g)$ at 900°C, the concentrations are: $[NH_3] = 0.12$ M, $[O_2] = 0.18$ M, $[NO] = 0.24$ M, $[H_2O] = 0.36$ M. What is $K_c$?
   Answer: 38 When you encounter equilibrium problems, you need to write the equilibrium constant expression using the balanced chemical equation, then substitute the given concentrations. For the reaction $4NH_3(g) + 5O_2(g) \rightleftharpoons 4NO(g) + 6H_2O(g)$, the equilibrium constant expression is: $K_c = \frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}$ Notice that products go in the numerator and reactants in the denominator, each raised to the power of their stoichiometric coefficients. Substituting the equilibrium concentrations: $K_c = \frac{(0.24)^4(0.36)^6}{(0.12)^4(0.18)^5}$ $K_c = \frac{(0.003317)(0.006047)}{(0.0000207)(0.00189)} = \frac{0.00002006}{0.0000000391} = 513$ Since 513 isn't among the choices A through D, the answer must be E (which typically indicates "none of the above" or a calculation outside the given range). Choice A (0.63) likely results from incorrectly placing products in the denominator and reactants in the numerator. Choice B (2.4) might come from using incorrect exponents or calculation errors. Choice C (6.9) and D (15) represent other computational mistakes, possibly from confusing concentration values or arithmetic errors in the complex calculation. Remember: always double-check your $K_c$ expression setup before calculating. Products over reactants, each raised to their stoichiometric coefficient - this is the most common source of error in equilibrium problems.
2. At 298 K, the reaction $2NO_2(g) \rightleftharpoons N_2O_4(g)$ reaches equilibrium with the following partial pressures: $P_{NO_2} = 0.45$ atm and $P_{N_2O_4} = 0.18$ atm. What is the value of $K_p$ for this reaction?
   Answer: 0.89 When you encounter equilibrium problems involving gas-phase reactions, you need to write the equilibrium expression using partial pressures and calculate $K_p$. For any reaction, $K_p$ equals the partial pressures of products raised to their stoichiometric coefficients, divided by the partial pressures of reactants raised to their coefficients. For this reaction, $2NO_2(g) \rightleftharpoons N_2O_4(g)$, the equilibrium expression is: $K_p = \frac{P_{N_2O_4}}{(P_{NO_2})^2}$ Notice that $NO_2$ is squared because its coefficient is 2, while $N_2O_4$ has an implied coefficient of 1. Substituting the given values: $K_p = \frac{0.18}{(0.45)^2} = \frac{0.18}{0.2025} = 0.89$ This confirms answer choice B is correct. Let's examine why the other options are wrong. Choice A (0.40) results from incorrectly calculating $(0.45)^2 = 0.45$, forgetting to square the $NO_2$ partial pressure. Choice C (1.1) comes from flipping the equilibrium expression, calculating $\frac{(0.45)^2}{0.18}$. Choice D (2.5) represents the reciprocal of the correct answer, which would be $K_p$ for the reverse reaction. Study tip: Always write the equilibrium expression first, paying careful attention to stoichiometric coefficients as exponents. Double-check that products are in the numerator and reactants in the denominator, and verify your arithmetic by ensuring the units work out correctly.
3. The equilibrium constant $K_c = 2.4 \times 10^{-3}$ for the reaction $2NOBr(g) \rightleftharpoons 2NO(g) + Br_2(g)$ at 25°C. What is $K_c$ for the reverse reaction $2NO(g) + Br_2(g) \rightleftharpoons 2NOBr(g)$?
   Answer: $4.2 \times 10^{2}$ When you encounter equilibrium constant problems involving reverse reactions, remember that the equilibrium constant for a reverse reaction is simply the reciprocal of the original constant. For the forward reaction $2NOBr(g) \rightleftharpoons 2NO(g) + Br_2(g)$, we have $K_c = 2.4 \times 10^{-3}$. The equilibrium expression is: $K_{c(forward)} = \frac{[NO]^2[Br_2]}{[NOBr]^2}$ For the reverse reaction $2NO(g) + Br_2(g) \rightleftharpoons 2NOBr(g)$, the equilibrium expression becomes: $K_{c(reverse)} = \frac{[NOBr]^2}{[NO]^2[Br_2]}$ Notice this is exactly the inverse of the forward expression, so $K_{c(reverse)} = \frac{1}{K_{c(forward)}} = \frac{1}{2.4 \times 10^{-3}} = 4.2 \times 10^{2}$. Looking at the wrong answers: Choice A ($2.4 \times 10^{-3}$) incorrectly assumes the equilibrium constant doesn't change when the reaction is reversed. Choice B ($4.9 \times 10^{-2}$) might result from taking the square root instead of the reciprocal. Choice C ($2.0 \times 10^{1}$) could come from calculation errors or misapplying mathematical operations. The correct answer is D: $4.2 \times 10^{2}$. Study tip: Always remember the reciprocal relationship for reverse reactions: $K_{reverse} = \frac{1}{K_{forward}}$. This is one of the most tested relationships in equilibrium problems, so master this concept early.
4. For the equilibrium $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ at 800°C, the equilibrium partial pressure of $CO_2$ is 0.23 atm. What is the value of $K_p$?
   Answer: 0.23 When you encounter equilibrium expressions involving solids and gases, remember that only gases and aqueous species appear in the equilibrium constant expression. Solids and pure liquids are omitted because their concentrations remain essentially constant. For this equilibrium, $K_p$ includes only the gaseous product $CO_2$. Since there are no gaseous reactants, the expression becomes simply: $K_p = P_{CO_2}$ Given that the equilibrium partial pressure of $CO_2$ is 0.23 atm, $K_p = 0.23$, making choice A correct. Let's examine why the other options are wrong. Choice B (0.48) might result from incorrectly squaring the pressure or applying some other mathematical manipulation that isn't warranted here. Choice C (1.0) could come from mistakenly thinking that equilibrium constants are always unity or from confusion about standard conditions. Choice D (4.3) might arise from taking the reciprocal (1/0.23 ≈ 4.3) or from incorrectly including the solid species in the calculation. The key insight is recognizing that this is a heterogeneous equilibrium where the equilibrium expression dramatically simplifies. Since $CaCO_3$ and $CaO$ are both solids, they don't appear in the $K_p$ expression at all. Study tip: For heterogeneous equilibria, always write out the $K_p$ expression first, excluding all solids and pure liquids. This prevents you from overcomplicating the calculation and helps you spot when the answer is surprisingly simple.
5. At 900 K, $K_c = 0.83$ for the reaction $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$. If a reaction mixture contains $[SO_2] = 0.40$ M, $[O_2] = 0.20$ M, and $[SO_3] = 0.60$ M, what can be concluded about the reaction?
   Answer: The reaction will proceed in reverse since Q > K When you encounter a problem asking whether a reaction will proceed forward or reverse, you need to compare the reaction quotient (Q) to the equilibrium constant (K). The reaction quotient uses the same expression as K but with current concentrations instead of equilibrium concentrations. For this reaction, $Q_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$. Substituting the given concentrations: $Q_c = \frac{(0.60)^2}{(0.40)^2(0.20)} = \frac{0.36}{0.032} = 11.25$ Since $Q_c = 11.25$ and $K_c = 0.83$, we have $Q_c > K_c$. When Q > K, the reaction must shift left (reverse direction) to reach equilibrium, consuming products and forming reactants until Q equals K. Looking at the wrong answers: Answer A is incorrect because Q (11.25) clearly does not equal K (0.83). Answer B represents a common mistake—when Q > K, the reaction goes reverse, not forward. If Q < K, then the reaction would proceed forward. Answer D makes no sense because Q has a definite numerical value (11.25), not zero, and reactions don't simply "not proceed." Remember this pattern: Q < K means forward reaction, Q > K means reverse reaction, and Q = K means equilibrium. Always calculate Q first, then compare it to the given K value to predict the reaction direction. This relationship is fundamental to understanding chemical equilibrium.
6. For the equilibrium $2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g)$, $K_c = 8.5 \times 10^{-3}$ at 1000 K. A reaction mixture at this temperature has $[SO_3] = 0.45$ M, $[SO_2] = 0.15$ M, and $[O_2] = 0.080$ M. What is the reaction quotient Q?
   Answer: $5.9 \times 10^{-3}$ When you encounter equilibrium problems, remember that the reaction quotient Q uses the same mathematical form as the equilibrium constant Kc, but with current concentrations rather than equilibrium concentrations. This allows you to determine which direction the reaction will proceed. For the reaction $2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g)$, the expression is: $Q = \frac{[SO_2]^2[O_2]}{[SO_3]^2}$ Notice that coefficients become exponents, and products go in the numerator while reactants go in the denominator. Substituting the given concentrations: $Q = \frac{(0.15)^2(0.080)}{(0.45)^2} = \frac{(0.0225)(0.080)}{0.2025} = \frac{0.0018}{0.2025} = 5.9 \times 10^{-3}$ This matches answer choice B. Let's examine why the other answers are incorrect: - Choice A ($2.4 \times 10^{-3}$) results from incorrectly using coefficients of 1 instead of 2 for the squared terms - Choice C ($1.2 \times 10^{-2}$) comes from forgetting to square the SO₃ concentration in the denominator - Choice D ($2.4 \times 10^{-2}$) occurs when you flip the expression, putting reactants over products Since Q < Kc ($5.9 \times 10^{-3} < 8.5 \times 10^{-3}$), this system will shift right toward products to reach equilibrium. Study tip: Always write the Q expression first, then carefully substitute concentrations. Double-check that you've applied the correct exponents based on the balanced equation coefficients.
7. At equilibrium, a 2.0 L container at 400°C contains 0.40 mol $PCl_5$, 0.20 mol $PCl_3$, and 0.60 mol $Cl_2$ for the reaction $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$. What is the value of $K_c$?
   Answer: 0.15 When you encounter an equilibrium problem asking for $K_c$, you need to write the equilibrium expression and substitute the equilibrium concentrations. The equilibrium constant $K_c$ equals the product of product concentrations divided by the product of reactant concentrations, each raised to their stoichiometric coefficients. For $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$, the equilibrium expression is: $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}$ First, convert moles to molarity by dividing by the 2.0 L volume: - $[PCl_5] = \frac{0.40 \text{ mol}}{2.0 \text{ L}} = 0.20 \text{ M}$ - $[PCl_3] = \frac{0.20 \text{ mol}}{2.0 \text{ L}} = 0.10 \text{ M}$ - $[Cl_2] = \frac{0.60 \text{ mol}}{2.0 \text{ L}} = 0.30 \text{ M}$ Substituting into the equilibrium expression: $K_c = \frac{(0.10)(0.30)}{(0.20)} = \frac{0.030}{0.20} = 0.15$ Answer A (0.15) is correct. Answer B (0.30) likely comes from using $[Cl_2]$ alone in the numerator and forgetting to multiply by $[PCl_3]$. Answer C (0.60) might result from using moles instead of molarity or making an arithmetic error. Answer D (1.8) could come from inverting the equilibrium expression, putting reactants over products. Always remember: convert to molarity first, write the correct equilibrium expression with products over reactants, and double-check your arithmetic. The equilibrium constant is dimensionless when coefficients are all 1.
8. For the equilibrium $CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)$ at 800 K, equal molar amounts of CO and $H_2O$ are mixed. At equilibrium, 60% of the CO has reacted. What is the equilibrium constant $K_c$?
   Answer: 2.25 When you encounter equilibrium problems with percentage conversion data, you need to set up an ICE table (Initial, Change, Equilibrium) to track concentrations and then apply the equilibrium constant expression. Let's assume you start with 1 mole each of CO and H₂O in a 1 L container. Since 60% of CO reacts, 0.6 mol of CO is consumed. The stoichiometry shows that equal moles of each reactant are consumed and each product is formed. Your ICE table looks like: - Initial: [CO] = 1 M, [H₂O] = 1 M, [CO₂] = 0 M, [H₂] = 0 M - Change: -0.6 M for both reactants, +0.6 M for both products - Equilibrium: [CO] = 0.4 M, [H₂O] = 0.4 M, [CO₂] = 0.6 M, [H₂] = 0.6 M The equilibrium constant expression is: $K_c = \frac{[CO_2][H_2]}{[CO][H_2O]} = \frac{(0.6)(0.6)}{(0.4)(0.4)} = \frac{0.36}{0.16} = 2.25$ This confirms answer B) 2.25 is correct. Answer A) 1.5 likely comes from incorrectly calculating the ratio of reacted to unreacted CO (0.6/0.4). Answer C) 3.0 might result from computational errors in the fraction. Answer D) 4.0 could come from mistakenly using (0.4)² in the numerator instead of denominator. Remember: always set up your ICE table systematically and double-check your equilibrium constant expression matches the balanced equation with products over reactants.
9. The following equilibrium concentrations were measured for the reaction $A + 2B \rightleftharpoons C + D$ at 298 K: $[A] = 0.15$ M, $[B] = 0.25$ M, $[C] = 0.45$ M, $[D] = 0.30$ M. What is the equilibrium constant $K_c$?
   Answer: 14 When you encounter equilibrium constant problems, you're applying the fundamental relationship between product and reactant concentrations at equilibrium. The equilibrium constant expression $K_c$ is written as products over reactants, with each concentration raised to the power of its stoichiometric coefficient. For the reaction $A + 2B \rightleftharpoons C + D$, the equilibrium constant expression is: $K_c = \frac{[C][D]}{[A][B]^2}$ Notice that [B] is squared because its coefficient is 2 in the balanced equation. Substituting the given equilibrium concentrations: $K_c = \frac{(0.45)(0.30)}{(0.15)(0.25)^2} = \frac{0.135}{(0.15)(0.0625)} = \frac{0.135}{0.009375} = 14.4$ This rounds to 14, confirming answer choice D. The wrong answers likely come from common calculation errors. Choice A (1.4) probably results from forgetting to square the [B] term or making an arithmetic mistake. Choice B (3.6) might come from incorrectly placing concentrations in the expression or computational errors. Choice C (7.2) could result from partially correct setup but flawed arithmetic. The key strategy here is methodical setup: always write the $K_c$ expression first, carefully noting stoichiometric coefficients as exponents. Double-check that products go in the numerator and reactants in the denominator. When calculating, pay special attention to exponents—this is where most errors occur in equilibrium constant problems.
10. Based on this information, what is the equilibrium constant $K_c$ for the dissociation reaction?
    Answer: 0.21 When you encounter equilibrium problems, you need to track how concentrations change from initial conditions to equilibrium, then apply the equilibrium expression. Start by setting up an ICE table (Initial, Change, Equilibrium). Initially, you have 2.0 mol $N_2O_4$ in 1.0 L, so [$N_2O_4$] = 2.0 M and [$NO_2$] = 0 M. The problem states that 0.30 mol of $N_2O_4$ dissociates, meaning the concentration decreases by 0.30 M. Since the stoichiometry shows 1 mol $N_2O_4$ produces 2 mol $NO_2$, the $NO_2$ concentration increases by 2(0.30) = 0.60 M. At equilibrium: [$N_2O_4$] = 2.0 - 0.30 = 1.7 M and [$NO_2$] = 0.60 M. The equilibrium expression is: $K_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(0.60)^2}{1.7} = \frac{0.36}{1.7} = 0.21$ Choice A (0.15) likely results from incorrectly using 0.30 instead of 0.60 for the $NO_2$ concentration, forgetting the 2:1 stoichiometry. Choice C (0.35) might come from using 1.0 M instead of 1.7 M for the $N_2O_4$ equilibrium concentration. Choice D (0.48) could result from calculation errors or mixing up the equilibrium concentrations. Remember: always account for stoichiometric coefficients when calculating concentration changes, and double-check that your equilibrium concentrations reflect the actual amounts remaining and formed.
11. For the reaction $2A(g) \rightleftharpoons B(g) + C(g)$, the equilibrium constant $K_c = 4.5 \times 10^{-3}$ at 400°C. What is $K_c$ for the reaction $B(g) + C(g) \rightleftharpoons 2A(g)$?
    Answer: $2.2 \times 10^{2}$ When you encounter equilibrium constant problems involving reaction reversals, remember that reversing a chemical equation creates a reciprocal relationship between the equilibrium constants. For the original reaction $2A(g) \rightleftharpoons B(g) + C(g)$, we have $K_c = 4.5 \times 10^{-3}$. The equilibrium constant expression is $K_c = \frac{[B][C]}{[A]^2}$. When we reverse this reaction to get $B(g) + C(g) \rightleftharpoons 2A(g)$, the equilibrium expression becomes $K_c' = \frac{[A]^2}{[B][C]}$. Notice this is exactly the reciprocal of the original expression. Therefore, $K_c' = \frac{1}{K_c} = \frac{1}{4.5 \times 10^{-3}} = 222 = 2.2 \times 10^{2}$. Choice A ($4.5 \times 10^{-3}$) incorrectly assumes the equilibrium constant doesn't change when you reverse the reaction. Choice B ($2.1 \times 10^{-2}$) appears to be $\sqrt{4.5 \times 10^{-3}} \times 10$, suggesting confusion about mathematical operations on equilibrium constants. Choice C ($6.7 \times 10^{-2}$) might result from incorrectly calculating $\sqrt{4.5 \times 10^{-3}}$, which would apply if the reaction were halved rather than reversed. The key principle to remember: when you reverse a reaction, take the reciprocal of the equilibrium constant. When you multiply coefficients by a factor, raise the equilibrium constant to that power. These mathematical relationships directly reflect how the concentration ratios change with the reaction manipulation.
12. At 1000 K, the equilibrium $C(s) + CO_2(g) \rightleftharpoons 2CO(g)$ has $K_p = 1.7$. If the equilibrium partial pressure of $CO_2$ is 0.50 atm, what is the equilibrium partial pressure of CO?
    Answer: 0.92 atm When you encounter equilibrium problems involving gases, you need to set up the equilibrium expression using partial pressures and solve for the unknown using the given $K_p$ value. For the reaction $C(s) + CO_2(g) \rightleftharpoons 2CO(g)$, the equilibrium expression is $K_p = \frac{P_{CO}^2}{P_{CO_2}}$. Notice that solid carbon doesn't appear in the expression since solids have constant activity. Given $K_p = 1.7$ and $P_{CO_2} = 0.50$ atm, you can substitute: $1.7 = \frac{P_{CO}^2}{0.50}$ Solving for $P_{CO}$: $P_{CO}^2 = 1.7 \times 0.50 = 0.85$, so $P_{CO} = \sqrt{0.85} = 0.92$ atm. Looking at the wrong answers: Choice A (0.58 atm) results from incorrectly taking the square root of $K_p$ instead of $K_p \times P_{CO_2}$. Choice B (0.85 atm) is the value of $P_{CO}^2$ before taking the square root—a common error when students forget that CO has a coefficient of 2 in the balanced equation. Choice D (1.3 atm) comes from incorrectly using $K_p = \frac{P_{CO}}{P_{CO_2}}$, ignoring the stoichiometric coefficient. The correct answer is C (0.92 atm). Study tip: Always write the $K_p$ expression first, paying careful attention to stoichiometric coefficients as exponents. Double-check your algebra, especially when taking square roots, as equilibrium problems often involve squared terms that students forget to properly handle.
13. For the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ at 731 K, equilibrium is established when $[H_2] = 0.025$ M, $[I_2] = 0.015$ M, and $[HI] = 0.12$ M. What is the equilibrium constant $K_c$ for this reaction?
    Answer: 38 When you encounter equilibrium problems, you're working with the equilibrium constant expression, which relates the concentrations of products and reactants at equilibrium. For any reaction, $K_c$ equals the product concentrations (raised to their stoichiometric coefficients) divided by the reactant concentrations (also raised to their coefficients). For this reaction, the equilibrium expression is: $K_c = \frac{[HI]^2}{[H_2][I_2]}$ Notice that HI has a coefficient of 2, so its concentration is squared. Substituting the given equilibrium concentrations: $K_c = \frac{(0.12)^2}{(0.025)(0.015)} = \frac{0.0144}{0.000375} = 38.4 \approx 38$ The answer is (B) 38. Looking at the wrong answers: (A) 3.1 likely results from forgetting to square the HI concentration—if you used $[HI]$ instead of $[HI]^2$, you'd get a much smaller value. (C) 64 might come from calculation errors, possibly rounding too early or misplacing decimal points. (D) 96 could result from incorrectly setting up the equilibrium expression, perhaps putting reactants in the numerator instead of the denominator. Study tip: Always write the equilibrium expression first, paying careful attention to stoichiometric coefficients—they become exponents. Double-check your setup before calculating, as the most common errors in equilibrium problems come from incorrect expressions rather than arithmetic mistakes.
14. At 700 K, $K_p = 0.76$ for $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$. What is $K_c$ at this temperature? (R = 0.0821 L·atm/mol·K)
    Answer: 44 When you encounter equilibrium problems involving both $K_p$ and $K_c$, you need to use the relationship $K_p = K_c(RT)^{\Delta n}$, where $\Delta n$ is the change in moles of gas from reactants to products. First, calculate $\Delta n$ by counting gas molecules on each side of the equation. For $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$, the reactants have 3 moles of gas (2 + 1) and the products have 2 moles of gas. Therefore, $\Delta n = 2 - 3 = -1$. Now solve for $K_c$: $0.76 = K_c(0.0821 × 700)^{-1}$ This gives us: $K_c = 0.76 × (0.0821 × 700) = 0.76 × 57.47 = 43.7$ Rounding to two significant figures gives $K_c = 44$, which is answer choice C. The wrong answers likely result from calculation errors or sign mistakes. Answer A (13) might come from incorrectly using $\Delta n = +1$ instead of $-1$. Answer B (24) could result from arithmetic errors in the multiplication. Answer D (67) might occur if you forget to multiply by the given $K_p$ value or make other computational mistakes. Remember that $\Delta n$ is always products minus reactants, and when $\Delta n$ is negative, $K_c$ will be larger than $K_p$. This makes sense because fewer gas molecules on the product side means higher concentration at equilibrium.
15. At 600°C, the reaction $H_2(g) + Br_2(g) \rightleftharpoons 2HBr(g)$ reaches equilibrium when $P_{H_2} = 0.30$ atm, $P_{Br_2} = 0.20$ atm, and $P_{HBr} = 1.8$ atm. Calculate $K_p$ for this reaction.
    Answer: 54 When you encounter an equilibrium problem asking for $K_p$, you need to write the equilibrium expression using partial pressures and substitute the given values. The equilibrium constant $K_p$ relates the partial pressures of products and reactants at equilibrium. For the reaction $H_2(g) + Br_2(g) \rightleftharpoons 2HBr(g)$, the equilibrium expression is: $K_p = \frac{(P_{HBr})^2}{(P_{H_2})(P_{Br_2})}$ Notice that $HBr$ has a coefficient of 2, so its partial pressure is squared in the expression. Substituting the equilibrium pressures: $K_p = \frac{(1.8)^2}{(0.30)(0.20)} = \frac{3.24}{0.060} = 54$ This confirms answer D is correct. Let's examine why the other answers are wrong. Answer A (3.0) would result from incorrectly writing $K_p = \frac{P_{HBr}}{(P_{H_2})(P_{Br_2})} = \frac{1.8}{0.060}$, forgetting to square the $HBr$ term. Answer B (9.0) comes from the calculation $(1.8)^2 = 3.24$, but then dividing by 0.36 instead of 0.060 – likely from adding the reactant pressures instead of multiplying them. Answer C (30) results from correctly squaring $HBr$ to get 3.24, but then dividing by 0.108 instead of 0.060, suggesting an arithmetic error. Always write the $K_p$ expression first, checking that exponents match the balanced equation coefficients. Then substitute carefully and double-check your arithmetic, especially when dealing with decimal multiplication and division.
16. At 1200 K, the reaction $C(s) + 2H_2(g) \rightleftharpoons CH_4(g)$ has an equilibrium mixture with $P_{H_2} = 0.50$ atm and $P_{CH_4} = 0.25$ atm. What is $K_p$ for this reaction?
    Answer: 1.0 When you encounter equilibrium problems, you need to write the equilibrium expression using partial pressures for gas-phase species, excluding pure solids and liquids since their activities equal 1. For this reaction, the equilibrium expression is: $K_p = \frac{P_{CH_4}}{P_{H_2}^2}$ Notice that carbon is a solid, so it doesn't appear in the expression. Only the gaseous species ($H_2$ and $CH_4$) are included. Substituting the given values: $K_p = \frac{0.25}{(0.50)^2} = \frac{0.25}{0.25} = 1.0$ Answer choice A (0.50) represents a common error where students might incorrectly calculate $\frac{P_{CH_4}}{P_{H_2}}$, forgetting to square the hydrogen pressure. Answer choice C (2.0) occurs when students flip the equilibrium expression, calculating $\frac{P_{H_2}^2}{P_{CH_4}}$ instead. Answer choice D (4.0) results from both flipping the expression AND forgetting to square the hydrogen pressure, giving $\frac{P_{H_2}}{P_{CH_4}}$. The correct answer is B (1.0). Remember: Always write the equilibrium expression exactly as the balanced equation appears, with products over reactants. Exclude pure solids and liquids from $K_p$ expressions, and don't forget that coefficients become exponents in the equilibrium expression. Double-check your calculation by ensuring you've squared any pressure terms that have a coefficient of 2 in the balanced equation.
17. The equilibrium constant for $A_2(g) + B_2(g) \rightleftharpoons 2AB(g)$ is $K_c = 50$ at 500 K. What is the equilibrium constant for $AB(g) \rightleftharpoons \frac{1}{2}A_2(g) + \frac{1}{2}B_2(g)$?
    Answer: 0.14 When you encounter questions about manipulating chemical equilibrium expressions, remember that equilibrium constants follow specific mathematical rules based on how you change the balanced equation. The original reaction has $K_c = 50$ for $A_2(g) + B_2(g) \rightleftharpoons 2AB(g)$. The target reaction is $AB(g) \rightleftharpoons \frac{1}{2}A_2(g) + \frac{1}{2}B_2(g)$. To transform the original equation into the target equation, you need two operations: first, reverse the reaction (flip reactants and products), then divide all coefficients by 2. When you reverse a reaction, the new equilibrium constant becomes $\frac{1}{K_c} = \frac{1}{50} = 0.020$. When you divide all coefficients by 2, you take the square root of the equilibrium constant: $\sqrt{0.020} = 0.14$. Looking at the wrong answers: Choice A (0.020) represents only reversing the reaction without accounting for halving the coefficients. Choice C (0.50) incorrectly applies $\frac{K_c}{100}$ instead of the proper mathematical operations. Choice D (7.1) appears to come from taking $\sqrt{50}$ without first reversing the reaction. The key strategy here is remembering the equilibrium constant manipulation rules: reversing a reaction gives you $\frac{1}{K}$, and multiplying coefficients by a factor $n$ raises $K$ to the $n$th power. Always apply these operations step-by-step in the order you manipulate the equation, and you'll avoid the common traps in equilibrium problems.
18. For the gas-phase reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, $K_p = 6.0 \times 10^{-2}$ at 500°C. What is the relationship between $K_c$ and $K_p$ for this reaction? (R = 0.0821 L·atm/mol·K)
    Answer: $K_c = K_p \times (RT)^2$ When you encounter gas-phase equilibrium problems asking about the relationship between $K_c$ and $K_p$, you need to consider how concentration and pressure units differ for gases. The key relationship is $K_p = K_c(RT)^{\Delta n}$, where $\Delta n$ is the change in moles of gas. For this reaction, calculate $\Delta n$ by subtracting moles of gaseous reactants from moles of gaseous products: $\Delta n = 2 - (1 + 3) = -2$. Therefore: $K_p = K_c(RT)^{-2}$. Rearranging to solve for $K_c$: $K_c = K_p(RT)^2$. Looking at the wrong answers: Choice A ($K_c = K_p$) would only be true if $\Delta n = 0$, meaning equal moles of gaseous reactants and products. That's not the case here. Choice C ($K_c = K_p(RT)^{-2}$) represents the original relationship before rearranging—this would give you $K_p$ in terms of $K_c$, not what we need. Choice D ($K_c = \frac{K_p}{RT}$) incorrectly uses $\Delta n = -1$, suggesting the person miscounted the moles or forgot to account for stoichiometric coefficients. Choice B correctly shows $K_c = K_p(RT)^2$, which accounts for the $\Delta n = -2$ and properly rearranges the fundamental relationship. Study tip: Always count moles of gaseous species carefully using stoichiometric coefficients, and remember that $\Delta n$ can be negative. The sign of $\Delta n$ determines whether $K_c$ is larger or smaller than $K_p$.
19. The equilibrium $2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g)$ has $K_c = 6.5 \times 10^{-6}$ at 500 K. What is $K_c$ for the reaction $NO(g) + \frac{1}{2}O_2(g) \rightleftharpoons NO_2(g)$?
    Answer: $3.9 \times 10^{2}$ When you encounter equilibrium constant problems involving related reactions, you need to understand how mathematical operations on chemical equations affect their equilibrium constants. The given reaction is $2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g)$ with $K_c = 6.5 \times 10^{-6}$. To find $K_c$ for $NO(g) + \frac{1}{2}O_2(g) \rightleftharpoons NO_2(g)$, you need to perform two operations: reverse the reaction and divide by 2. First, reverse the original reaction: $2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)$. When you reverse a reaction, the new equilibrium constant equals the reciprocal of the original: $K_{reverse} = \frac{1}{6.5 \times 10^{-6}} = 1.54 \times 10^5$. Next, divide the entire equation by 2 to get the target reaction. When you divide a reaction by a factor, you raise the equilibrium constant to the power of $\frac{1}{\text{factor}}$: $K_{final} = (1.54 \times 10^5)^{1/2} = 392 \approx 3.9 \times 10^2$. This matches answer C. Answer A ($2.5 \times 10^{-3}$) results from only taking the square root without reversing the reaction. Answer B ($1.5 \times 10^{5}$) comes from reversing but forgetting to take the square root. Answer D ($6.2 \times 10^{3}$) likely stems from calculation errors in the square root step. Remember: reversing a reaction takes the reciprocal of $K_c$, while multiplying coefficients by a factor raises $K_c$ to that power. Always track both the direction and stoichiometry changes carefully.

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