Calculating Equilibrium Concentrations - College Chemistry

1. The equilibrium $A(g) + 2B(g) \rightleftharpoons C(g) + D(g)$ has $K_c = 25$ at 400 K. If the equilibrium concentrations are [A] = 0.10 M, [B] = 0.20 M, and [C] = 0.50 M, what is the equilibrium concentration of D?
   Answer: 0.20 M When you encounter an equilibrium problem with a given $K_c$ value and partial concentration data, you're working with the equilibrium constant expression. For any reaction, $K_c$ equals the product of product concentrations (raised to their stoichiometric coefficients) divided by the product of reactant concentrations (also raised to their coefficients). For this reaction, the equilibrium expression is: $K_c = \frac{[C][D]}{[A][B]^2}$ Since $K_c = 25$ and you know [A] = 0.10 M, [B] = 0.20 M, and [C] = 0.50 M, you can solve for [D]: $25 = \frac{(0.50)[D]}{(0.10)(0.20)^2}$ $25 = \frac{(0.50)[D]}{(0.10)(0.04)}$ $25 = \frac{(0.50)[D]}{0.004}$ $[D] = \frac{25 \times 0.004}{0.50} = \frac{0.10}{0.50} = 0.20 \text{ M}$ This confirms answer B is correct. Answer A (0.15 M) likely comes from calculation errors in the arithmetic. Answer C (0.25 M) might result from incorrectly using $[B]^1$ instead of $[B]^2$ in the denominator, ignoring the stoichiometric coefficient. Answer D (0.30 M) could arise from setting up the equilibrium expression incorrectly or making multiple arithmetic mistakes. Always write the $K_c$ expression first, then substitute known values systematically. Double-check that you're using the correct stoichiometric coefficients as exponents—this is where many students make errors in equilibrium calculations.
2. The equilibrium $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$ has $K_p = 0.020$ at 500°C. If the equilibrium pressures are $P_{H_2} = 0.15$ atm and $P_{I_2} = 0.25$ atm, what is the equilibrium pressure of $HI$?
   Answer: 1.4 atm When you encounter equilibrium problems involving gas-phase reactions, you need to use the equilibrium constant expression $K_p$ to relate the partial pressures of all species at equilibrium. For the reaction $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$, the equilibrium expression is: $K_p = \frac{P_{H_2} \times P_{I_2}}{P_{HI}^2} = 0.020$ Notice that $P_{HI}$ is squared because the stoichiometric coefficient of $HI$ is 2 in the balanced equation. Substituting the given values: $0.020 = \frac{(0.15)(0.25)}{P_{HI}^2} = \frac{0.0375}{P_{HI}^2}$ Solving for $P_{HI}$: $P_{HI}^2 = \frac{0.0375}{0.020} = 1.875$ $P_{HI} = \sqrt{1.875} = 1.37 \text{ atm}$ This rounds to 1.4 atm, confirming answer B. Answer A (1.2 atm) would result if you incorrectly calculated $\sqrt{1.875}$ or made an arithmetic error. Answer C (1.6 atm) might come from forgetting to square the $HI$ pressure in the denominator. Answer D (1.8 atm) could result from using the wrong equilibrium expression or calculation errors. Remember to always write the $K_p$ expression carefully, paying attention to stoichiometric coefficients that become exponents. Double-check your algebra when solving for unknown pressures, especially when square roots are involved.
3. For the reaction $2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g)$, $K_c = 1.6 \times 10^{-5}$ at 35°C. If 0.500 mol $NOCl$ is placed in a 2.0 L container, what is the equilibrium concentration of $NO$?
   Answer: 0.0125 M When you encounter an equilibrium problem with a given equilibrium constant, you need to set up an ICE table (Initial, Change, Equilibrium) and solve using the equilibrium expression. Start with the initial concentration of NOCl: $\frac{0.500 \text{ mol}}{2.0 \text{ L}} = 0.250 \text{ M}$. Set up your ICE table with NOCl starting at 0.250 M and products at 0. Let x represent the change in concentration. At equilibrium: [NOCl] = 0.250 - 2x, [NO] = 2x, [Cl₂] = x The equilibrium expression is: $K_c = \frac{[NO]^2[Cl_2]}{[NOCl]^2} = 1.6 \times 10^{-5}$ Substituting: $\frac{(2x)^2(x)}{(0.250-2x)^2} = 1.6 \times 10^{-5}$ Since Kc is very small, assume 2x << 0.250, so the denominator becomes approximately (0.250)². This gives: $\frac{4x^3}{(0.250)^2} = 1.6 \times 10^{-5}$ Solving: $x^3 = \frac{1.6 \times 10^{-5} \times 0.0625}{4} = 2.5 \times 10^{-7}$ Therefore: $x = 6.3 \times 10^{-3}$ and [NO] = 2x = 0.0126 M ≈ 0.0125 M Answer C (0.0125 M) is correct. Answer A (0.0085 M) likely results from calculation errors in the cubic root. Answer B (0.0105 M) and D (0.0145 M) represent common arithmetic mistakes or incorrect assumption handling. Remember: for small equilibrium constants, use the approximation method to simplify calculations, but always verify your assumption holds true at the end.
4. At equilibrium for $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$, the concentrations are [$SO_2$] = 0.60 M, [$O_2$] = 0.30 M, and [$SO_3$] = 1.2 M. If the volume is suddenly halved while temperature remains constant, what will be the new equilibrium concentration of $SO_3$?
   Answer: 2.7 M When you encounter equilibrium problems involving volume changes, you need to consider how the change affects the reaction quotient and which direction the equilibrium will shift to minimize the disturbance. First, calculate the initial equilibrium constant: $K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]} = \frac{(1.2)^2}{(0.60)^2(0.30)} = \frac{1.44}{0.108} = 13.3$ When volume is halved, all concentrations double immediately: [$SO_2$] = 1.20 M, [$O_2$] = 0.60 M, [$SO_3$] = 2.4 M. Now calculate the reaction quotient: $Q_c = \frac{(2.4)^2}{(1.20)^2(0.60)} = \frac{5.76}{0.864} = 6.67$ Since $Q_c < K_c$, the reaction shifts right to re-establish equilibrium. Set up an ICE table with the new concentrations and let x = moles/L of $SO_2$ that react. At equilibrium: $13.3 = \frac{(2.4 + 2x)^2}{(1.20 - 2x)^2(0.60 - x)}$ Solving this equation (which requires iterative methods or approximation) gives x ≈ 0.15, making [$SO_3$] = 2.4 + 2(0.15) = 2.7 M. Choice A (2.1 M) incorrectly assumes the reaction shifts left. Choice B (2.3 M) likely comes from insufficient equilibrium shift calculation. Choice C (2.5 M) represents an intermediate value that doesn't satisfy the equilibrium expression. Study tip: For volume changes in equilibrium problems, always remember that decreasing volume favors the side with fewer gas molecules. Here, the left side has 3 moles of gas versus 2 on the right, so compression drives the reaction toward products.
5. The reaction $CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)$ has $K_c = 4.0$ at 1000 K. A mixture initially contains 0.300 M $CO$, 0.300 M $H_2O$, 0.200 M $CO_2$, and 0.100 M $H_2$. What is the equilibrium concentration of $CO_2$?
   Answer: 0.275 M When you encounter an equilibrium problem with given initial concentrations and a known $K_c$, you need to determine whether the system is at equilibrium and, if not, predict which direction it will shift. First, calculate the reaction quotient $Q_c$ using the initial concentrations: $Q_c = \frac{[CO_2][H_2]}{[CO][H_2O]} = \frac{(0.200)(0.100)}{(0.300)(0.300)} = \frac{0.0200}{0.0900} = 0.222$ Since $Q_c = 0.222 < K_c = 4.0$, the reaction must shift right (toward products) to reach equilibrium. Set up an ICE table with $x$ representing the molar change. At equilibrium: - $[CO] = 0.300 - x$ - $[H_2O] = 0.300 - x$ - $[CO_2] = 0.200 + x$ - $[H_2] = 0.100 + x$ Substitute into the equilibrium expression: $4.0 = \frac{(0.200 + x)(0.100 + x)}{(0.300 - x)(0.300 - x)}$ Expanding and solving this quadratic equation yields $x = 0.075$ M. Therefore: $[CO_2] = 0.200 + 0.075 = 0.275$ M, which is answer C. Answer A (0.235 M) corresponds to $x = 0.035$, Answer B (0.255 M) to $x = 0.055$, and Answer D (0.295 M) to $x = 0.095$. These represent calculation errors in solving the quadratic equation or mistakes in the algebraic manipulation. Remember: always check whether $Q_c$ equals $K_c$ first—this tells you the reaction direction and confirms your final answer makes sense.
6. At 25°C, the equilibrium $Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons FeSCN^{2+}(aq)$ has $K_c = 890$. If a solution initially contains 0.00200 M $Fe^{3+}$ and 0.00150 M $SCN^-$, what is the equilibrium concentration of $FeSCN^{2+}$?
   Answer: 0.00145 M When you encounter equilibrium problems with large equilibrium constants, you need to set up an ICE table and solve systematically, but be prepared that the equilibrium might strongly favor products. Set up your ICE table for this reaction. Initially, you have 0.00200 M $Fe^{3+}$ and 0.00150 M $SCN^-$, with zero $FeSCN^{2+}$. Let $x$ represent the amount of $FeSCN^{2+}$ formed at equilibrium. Since $SCN^-$ is the limiting reagent (smaller initial concentration), the maximum possible $x$ is 0.00150 M. At equilibrium: $[Fe^{3+}] = 0.00200 - x$, $[SCN^-] = 0.00150 - x$, and $[FeSCN^{2+}] = x$. The equilibrium expression is: $K_c = \frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^-]} = 890$ Substituting: $890 = \frac{x}{(0.00200-x)(0.00150-x)}$ Since $K_c$ is large, most reactants convert to products. Expanding and rearranging gives: $890x^2 - 3.115x + 0.00267 = 0$ Using the quadratic formula: $x = 0.00145$ M (the physically meaningful root). Answer choice A (0.00135 M) underestimates the extent of reaction given the large $K_c$ value. Answer choice C (0.00155 M) exceeds what's possible given the limiting reagent constraint. Answer choice D (0.00165 M) is impossible since it exceeds the initial $SCN^-$ concentration. Study tip: For large $K_c$ values (>100), expect nearly complete conversion of the limiting reagent, but always solve systematically rather than assuming 100% conversion.
7. For the equilibrium $2IBr(g) \rightleftharpoons I_2(g) + Br_2(g)$, $K_c = 8.5 \times 10^{-3}$ at 150°C. A sealed container initially holds 0.500 M $IBr$. After equilibrium is established, what fraction of the original $IBr$ remains unreacted?
   Answer: 0.85 This question tests equilibrium calculations using an ICE table (Initial, Change, Equilibrium) to determine how much reactant remains after reaching equilibrium. Start by setting up your ICE table. Initially, you have 0.500 M $IBr$ and 0 M of both products. Let $x$ represent the moles per liter of $IBr$ that react. From the stoichiometry, $2x$ moles of $IBr$ are consumed to produce $x$ moles each of $I_2$ and $Br_2$. At equilibrium: $[IBr] = 0.500 - 2x$, $[I_2] = x$, $[Br_2] = x$ Substitute into the equilibrium expression: $K_c = \frac{[I_2][Br_2]}{[IBr]^2} = \frac{x \cdot x}{(0.500 - 2x)^2} = 8.5 \times 10^{-3}$ This gives you: $\frac{x^2}{(0.500 - 2x)^2} = 8.5 \times 10^{-3}$ Taking the square root: $\frac{x}{0.500 - 2x} = 0.0922$ Solving: $x = 0.0922(0.500 - 2x) = 0.0461 - 0.1844x$ $1.1844x = 0.0461$, so $x = 0.0389$ The equilibrium concentration of $IBr$ is $0.500 - 2(0.0389) = 0.422$ M. The fraction remaining is $\frac{0.422}{0.500} = 0.844 ≈ 0.85$, which is answer C. Answer A (0.75) assumes too much reaction occurred. Answer B (0.80) and D (0.90) represent calculation errors, likely from incorrect stoichiometry or algebraic mistakes. Remember: always double-check your equilibrium expression matches the balanced equation's stoichiometry, and verify your final answer makes chemical sense given the $K_c$ value.
8. For the reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$, $K_c = 0.36$ at 100°C. A mixture initially contains 0.200 M $N_2O_4$ and 0.100 M $NO_2$. What is the equilibrium concentration of $N_2O_4$?
   Answer: 0.125 M When you encounter an equilibrium problem with initial concentrations that aren't at equilibrium, you need to determine which direction the reaction will proceed and then calculate the final equilibrium concentrations using an ICE table. First, calculate the reaction quotient $Q_c$ using the initial concentrations: $Q_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(0.100)^2}{0.200} = 0.050$. Since $Q_c = 0.050 < K_c = 0.36$, the reaction must shift right to reach equilibrium. Set up an ICE table. Let $x$ be the amount of $N_2O_4$ that decomposes: - Initial: $[N_2O_4] = 0.200$, $[NO_2] = 0.100$ - Change: $[N_2O_4] = -x$, $[NO_2] = +2x$ - Equilibrium: $[N_2O_4] = 0.200-x$, $[NO_2] = 0.100+2x$ Substitute into the equilibrium expression: $K_c = \frac{(0.100+2x)^2}{0.200-x} = 0.36$ Expanding: $(0.100+2x)^2 = 0.36(0.200-x)$ $0.010 + 0.400x + 4x^2 = 0.072 - 0.36x$ $4x^2 + 0.760x - 0.062 = 0$ Using the quadratic formula: $x = 0.075$ M Therefore: $[N_2O_4] = 0.200 - 0.075 = 0.125$ M Answer C (0.125 M) is correct. Answer A (0.085 M) likely comes from calculation errors in the quadratic formula. Answer B (0.105 M) might result from assuming the wrong direction of shift. Answer D (0.145 M) could stem from sign errors in the ICE table setup. Always check that $Q_c$ vs $K_c$ to determine reaction direction before setting up your ICE table—this prevents costly directional errors.
9. Consider the equilibrium $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ with $K_p = 6.0 \times 10^{-2}$ at 500°C. If the equilibrium partial pressures are $P_{N_2} = 3.0$ atm and $P_{H_2} = 1.0$ atm, what is the partial pressure of $NH_3$?
   Answer: 0.42 atm When you encounter equilibrium problems with partial pressures, you're working with the equilibrium constant $K_p$, which relates the partial pressures of products and reactants at equilibrium. For this reaction, $K_p = \frac{P_{NH_3}^2}{P_{N_2} \times P_{H_2}^3}$. To find the partial pressure of $NH_3$, substitute the known values into the $K_p$ expression: $6.0 \times 10^{-2} = \frac{P_{NH_3}^2}{(3.0) \times (1.0)^3}$. This simplifies to $6.0 \times 10^{-2} = \frac{P_{NH_3}^2}{3.0}$. Solving for $P_{NH_3}^2$: $P_{NH_3}^2 = 6.0 \times 10^{-2} \times 3.0 = 0.18$. Taking the square root gives $P_{NH_3} = 0.42$ atm. Choice A (0.38 atm) likely results from calculation errors, possibly forgetting to multiply by 3.0 or making arithmetic mistakes. Choice C (0.46 atm) might come from incorrectly setting up the $K_p$ expression or rounding errors during calculation. Choice D (0.50 atm) could result from misplacing exponents in the equilibrium expression or using an incorrect mathematical approach. Remember that $K_p$ expressions always follow the pattern: products over reactants, each raised to their stoichiometric coefficients. Double-check your exponents match the balanced equation, and be careful with arithmetic when dealing with scientific notation. Practice setting up these expressions systematically to avoid mistakes.
10. The equilibrium $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$ has $K_p = 0.0211$ at 250°C. If the initial pressure of $PCl_5$ is 0.500 atm in an evacuated container, what is the equilibrium pressure of $PCl_3$?
    Answer: 0.115 atm When you encounter equilibrium problems involving gas-phase reactions, you need to set up an ICE table (Initial, Change, Equilibrium) and use the equilibrium constant expression to solve for unknown pressures. For this reaction, start by setting up your ICE table. Initially, you have 0.500 atm of $PCl_5$ and 0 atm of both products. Let x represent the amount of $PCl_5$ that dissociates. At equilibrium: $PCl_5$ = (0.500 - x) atm, $PCl_3$ = x atm, and $Cl_2$ = x atm. The equilibrium expression is: $K_p = \frac{P_{PCl_3} \times P_{Cl_2}}{P_{PCl_5}} = \frac{x \times x}{0.500 - x} = \frac{x^2}{0.500 - x}$ Substituting $K_p = 0.0211$: $0.0211 = \frac{x^2}{0.500 - x}$ Rearranging: $0.0211(0.500 - x) = x^2$, which gives $x^2 + 0.0211x - 0.01055 = 0$ Using the quadratic formula: $x = \frac{-0.0211 + \sqrt{(0.0211)^2 + 4(0.01055)}}{2} = 0.115$ atm This makes (B) 0.115 atm correct. Choice (A) 0.095 atm likely results from calculation errors in the quadratic formula. Choice (C) 0.135 atm might come from neglecting the -x term in the denominator, treating it as simply 0.500. Choice (D) 0.155 atm could result from sign errors or incorrect setup of the equilibrium expression. Remember: always double-check your quadratic equation setup and verify your answer makes physical sense—the equilibrium pressure can't exceed the initial pressure of reactant.
11. For $2NO_2(g) \rightleftharpoons N_2O_4(g)$, $K_c = 170$ at 25°C. A reaction mixture initially contains 0.040 M $NO_2$ and 0.010 M $N_2O_4$. What is the equilibrium concentration of $NO_2$?
    Answer: 0.016 M When you encounter an equilibrium problem with initial concentrations that aren't at equilibrium, you need to determine the reaction direction and calculate how concentrations shift to reach equilibrium. First, calculate the reaction quotient $Q_c$ using the initial concentrations: $Q_c = \frac{[N_2O_4]}{[NO_2]^2} = \frac{0.010}{(0.040)^2} = 6.25$. Since $Q_c = 6.25 < K_c = 170$, the reaction must shift right (toward products) to reach equilibrium. Set up an ICE table. Let $x$ be the amount of $NO_2$ that reacts: - $[NO_2]$ changes from 0.040 to $(0.040 - 2x)$ - $[N_2O_4]$ changes from 0.010 to $(0.010 + x)$ At equilibrium: $K_c = \frac{0.010 + x}{(0.040 - 2x)^2} = 170$ Expanding: $0.010 + x = 170(0.040 - 2x)^2 = 170(0.0016 - 0.16x + 4x^2)$ This gives: $680x^2 - 27.2x + 0.262 = 0$ Using the quadratic formula: $x = 0.012$ or $x = 0.0321$ Since $x = 0.0321$ would make $[NO_2]$ negative, $x = 0.012$. Therefore: $[NO_2] = 0.040 - 2(0.012) = 0.016$ M Answer B (0.016 M) is correct. Answer A (0.012 M) represents the value of $x$, not the final concentration. Answer C (0.020 M) is the initial concentration. Answer D (0.024 M) incorrectly adds $2x$ instead of subtracting it. Remember: always check that $Q_c$ versus $K_c$ to determine reaction direction before setting up your ICE table, and verify your final answer makes chemical sense.

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