Representations of Equilibrium - College Chemistry

1. Consider the equilibrium $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$ with $K_{sp} = 5.6 \times 10^{-12}$. A student writes the equilibrium expression as $K_{sp} = \frac{[Mg^{2+}][OH^-]^2}{[Mg(OH)_2]}$ and argues that this accounts for the limited solubility of the solid. How should this argument be evaluated?
   Answer: The argument is incorrect because pure solids are always omitted from equilibrium expressions When you encounter equilibrium expressions involving solids and aqueous ions, remember that the physical state of each species determines whether it appears in the equilibrium expression. The correct equilibrium expression for this dissolution reaction is simply $K_{sp} = [Mg^{2+}][OH^-]^2$. Pure solids (and pure liquids) are always omitted from equilibrium expressions because their concentrations remain essentially constant regardless of how much solid is present. The "concentration" of a pure solid is determined by its density, which doesn't change whether you have 1 gram or 100 grams of $Mg(OH)_2$. This constant value gets incorporated into the equilibrium constant itself. Choice A incorrectly suggests that limited solubility means the solid concentration varies—but the solid's concentration stays constant even as it dissolves. Choice C mentions surface area, but while surface area affects the rate of dissolution, it doesn't belong in the equilibrium expression itself. Choice D suggests the expression is correct for saturated solutions, but equilibrium expressions have the same form regardless of whether equilibrium has been reached. Choice B correctly identifies that pure solids are always excluded from equilibrium expressions, making the student's inclusion of $[Mg(OH)_2]$ in the denominator fundamentally wrong. Study tip: For any heterogeneous equilibrium, only include aqueous and gaseous species in your equilibrium expression. Pure solids and liquids always get omitted—their "concentrations" are built into the K value itself.
2. Consider the acid ionization equilibrium $HC_2H_3O_2(aq) \rightleftharpoons H^+(aq) + C_2H_3O_2^-(aq)$ with $K_a = 1.8 \times 10^{-5}$. Which expression correctly represents this equilibrium constant?
   Answer: $K_a = \frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$ When you encounter acid ionization equilibrium problems, you need to write the equilibrium constant expression using the standard rule: products over reactants, with each concentration raised to its stoichiometric coefficient. For the given equilibrium $HC_2H_3O_2(aq) \rightleftharpoons H^+(aq) + C_2H_3O_2^-(aq)$, the correct expression is $K_a = \frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$. The products ($H^+$ and $C_2H_3O_2^-$) go in the numerator, and the reactant ($HC_2H_3O_2$) goes in the denominator. Notice that water doesn't appear in this expression because it's the solvent in aqueous solutions - its concentration remains essentially constant and is incorporated into the equilibrium constant value. Looking at the wrong answers: Choice A incorrectly includes $[H_2O]$ in the denominator. While water is technically a product when you write the complete reaction, we don't include solvent concentrations in aqueous equilibrium expressions. Choice C is missing the denominator entirely - this would suggest that the equilibrium doesn't depend on the concentration of the acid, which is clearly wrong. Choice D has the expression completely inverted, putting reactants over products, which would give you $1/K_a$ instead of $K_a$. Remember this key pattern: for any equilibrium expression, always write products over reactants, and exclude pure liquids and solids (including water as solvent) from the expression. This rule applies to all equilibrium constants, whether $K_a$, $K_b$, or $K_{eq}$.
3. For the reaction $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$, the equilibrium constant expression is written as $K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$. A student claims this expression is incorrect because it doesn't account for the different phases present. Which statement best evaluates this claim?
   Answer: The claim is incorrect; all species are in the gas phase, so the expression is valid for $K_c$ When you encounter equilibrium constant expressions, the key consideration is whether all species are in the same phase and how that affects the mathematical form of the expression. The given expression $K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$ is perfectly valid because all reactants and products are in the gas phase. The equilibrium constant expression follows the standard form: products in the numerator, reactants in the denominator, each raised to the power of their stoichiometric coefficients. The fact that these are all gases doesn't invalidate using concentrations for $K_c$. Let's examine why the other options are incorrect. Option A suggests that gases must use partial pressures ($K_p$), but this is misleading—you can express equilibrium constants for gas-phase reactions using either concentrations ($K_c$) or partial pressures ($K_p$). Both are valid, though they have different numerical values related by $K_p = K_c(RT)^{\Delta n}$. Option C mentions activity coefficients for non-ideal behavior, but standard equilibrium expressions assume ideal conditions unless specifically stated otherwise. Option D incorrectly claims that phase considerations never matter in equilibrium expressions, when in fact they're crucial for heterogeneous equilibria involving solids or liquids. The student's claim about "different phases" is the root of their confusion—there's only one phase (gas) present here. Study tip: Remember that $K_c$ and $K_p$ are both valid for gas-phase reactions. Phase considerations only become critical in heterogeneous equilibria where you exclude pure solids and liquids from the expression.
4. Consider an equilibrium system where $K_c = 2.5 \times 10^{-3}$ at 298 K. A student calculates the reaction quotient Q at a particular moment and finds $Q = 1.8 \times 10^{-4}$. Based on the relationship between Q and $K_c$, what can be concluded about this system?
   Answer: The forward reaction will proceed because $Q < K_c$, meaning more products must form When you encounter questions comparing reaction quotient Q to equilibrium constant $K_c$, you're dealing with Le Chatelier's principle and predicting reaction direction. The key is understanding that Q tells you the current state while $K_c$ tells you where the system wants to be. Since $Q = 1.8 \times 10^{-4}$ and $K_c = 2.5 \times 10^{-3}$, we have $Q < K_c$. This means the reaction quotient is smaller than it should be at equilibrium. Since Q has the same mathematical form as $K_c$ (products over reactants), a smaller Q indicates there are currently fewer products and more reactants than there should be at equilibrium. Therefore, the forward reaction must proceed to produce more products until Q equals $K_c$. This makes B correct. Option A is wrong because equilibrium isn't determined by whether values are "small" - it's determined by whether Q equals $K_c$. Option C misinterprets what $Q < K_c$ means; it actually indicates too few products, not too many, so the reverse reaction won't be favored. Option D incorrectly suggests temperature changes are needed. Temperature affects the value of $K_c$, but here we're simply observing a system approaching its existing equilibrium at constant temperature. Remember this pattern: $Q < K_c$ means forward reaction proceeds (need more products), $Q > K_c$ means reverse reaction proceeds (need fewer products), and $Q = K_c$ means equilibrium. The relative magnitudes, not absolute values, determine direction.
5. For the equilibrium $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$, the equilibrium constant $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$ at 25°C. A student writes the equilibrium expression as $K = \frac{[H^+][OH^-]}{[H_2O]}$. What error has the student made?
   Answer: The student included pure liquid water, which should be omitted from equilibrium expressions When writing equilibrium expressions, you need to understand which species to include and which to omit. The key principle is that pure solids and pure liquids are omitted from equilibrium expressions because their concentrations remain essentially constant during the reaction. In the autoionization of water, liquid water ($H_2O(l)$) is a pure liquid with a constant concentration of approximately 55.6 M at 25°C. Since this value doesn't change significantly during the equilibrium, it's incorporated into the equilibrium constant itself. This is why we write $K_w = [H^+][OH^-]$ rather than including water in the denominator. Choice B correctly identifies that the student included pure liquid water, which should be omitted from equilibrium expressions. The constant concentration of water is already "built into" the value of $K_w$. Choice A is incorrect because there's no such thing as an "autoionization coefficient" that belongs in equilibrium expressions. Choice C is wrong because while activities are more theoretically rigorous than concentrations, this isn't the error being tested here—the student's main mistake is including liquid water. Choice D is incorrect because the stoichiometry is already correctly represented; each ion has a coefficient of 1, so no squaring is needed. Remember this rule: pure solids and pure liquids are always omitted from equilibrium expressions. Only include aqueous solutions, gases, and species whose concentrations can actually change during the reaction.
6. For the equilibrium $2NO_2(g) \rightleftharpoons N_2O_4(g)$, $K_c = 4.6$ at 25°C. What is the correct expression for $K_c$ for the reverse reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$?
   Answer: $K_c = \frac{[NO_2]^2}{[N_2O_4]} = 0.22$ When you encounter equilibrium constant problems involving reverse reactions, remember that the equilibrium expression and its value both change in predictable ways. For the original reaction $2NO_2(g) \rightleftharpoons N_2O_4(g)$, the equilibrium expression is $K_c = \frac{[N_2O_4]}{[NO_2]^2} = 4.6$. When you reverse a reaction, two things happen: the equilibrium expression flips (products become reactants and vice versa), and the equilibrium constant becomes the reciprocal of the original value. For the reverse reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$, the correct expression is $K_c = \frac{[NO_2]^2}{[N_2O_4]}$, and the value becomes $\frac{1}{4.6} = 0.22$. This makes answer choice B correct. Let's examine why the other options fail: Choice A has the correct expression for the reverse reaction but incorrectly keeps the original K value of 4.6 instead of taking its reciprocal. Choice C makes the opposite error—it uses the original reaction's expression ($\frac{[N_2O_4]}{[NO_2]^2}$) but applies the reciprocal value. Choice D keeps both the original expression and original K value, showing no recognition that the reaction was reversed. Study tip: For reversed equilibrium reactions, remember "flip and flip"—flip the expression (invert the fraction) and flip the K value (take the reciprocal). Both must change together.
7. Consider the equilibrium $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$. Which expression correctly represents the equilibrium constant for this reaction?
   Answer: $K = [CO_2]$ When writing equilibrium expressions, you need to understand how the physical states of reactants and products affect the expression. The key principle is that pure solids and pure liquids are omitted from equilibrium expressions because their concentrations remain essentially constant during the reaction. For this equilibrium, $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$, you have two pure solids ($CaCO_3$ and $CaO$) and one gas ($CO_2$). Since the solids maintain constant "activity" (essentially constant concentration), they don't appear in the equilibrium expression. Only the gas phase species $CO_2$ has a concentration that can vary meaningfully, so $K = [CO_2]$, making B correct. Option A ($K = \frac{[CaO][CO_2]}{[CaCO_3]}$) incorrectly includes both solid phases, treating them as if they were aqueous solutions or gases. Option C ($K = \frac{[CO_2]}{[CaCO_3]}$) correctly excludes $CaO$ but wrongly includes the solid $CaCO_3$ in the denominator. Option D ($K = \frac{[CaO]}{[CaCO_3]}$) includes both solids while completely omitting the gas, which makes no chemical sense since $CO_2$ pressure is what actually determines the equilibrium position. Remember this pattern: when writing equilibrium expressions, automatically exclude pure solids and pure liquids, keeping only gases and aqueous species. This is one of the most commonly tested concepts in equilibrium chemistry, so practice identifying phases from chemical formulas and their state symbols.
8. For the equilibrium $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$, the equilibrium constant can be expressed as either $K_c$ or $K_p$. At any given temperature, what is the relationship between these two constants?
   Answer: $K_p = K_c$ because $\Delta n = 0$ When you encounter equilibrium constant questions involving both $K_c$ and $K_p$, you need to consider how the number of moles of gas changes during the reaction. The relationship between these constants is given by: $K_p = K_c(RT)^{\Delta n}$, where $\Delta n$ is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants). For this reaction, let's calculate $\Delta n$: We have 2 moles of HI(g) as products and 1 mole each of H₂(g) and I₂(g) as reactants. So $\Delta n = 2 - (1 + 1) = 0$. When $\Delta n = 0$, the equation becomes $K_p = K_c(RT)^0 = K_c(1) = K_c$. Therefore, $K_p = K_c$. Looking at the wrong answers: (A) incorrectly assumes that simply having gases involved makes $K_p > K_c$, but the relationship depends on $\Delta n$, not just the presence of gases. (C) suggests pressure effects make $K_p < K_c$, but again ignores that $\Delta n = 0$ eliminates any pressure-related differences between the constants. (D) incorrectly thinks the coefficient 2 in front of HI directly affects the relationship between $K_p$ and $K_c$, but coefficients are already accounted for in the equilibrium expressions themselves. The correct answer is (B): $K_p = K_c$ because $\Delta n = 0$. Study tip: Always calculate $\Delta n$ first when comparing $K_c$ and $K_p$. When $\Delta n = 0$, the constants are equal regardless of temperature or pressure.
9. For the coupled equilibria: $A(g) \rightleftharpoons B(g)$ with $K_1 = 2.0$ and $B(g) \rightleftharpoons C(g)$ with $K_2 = 3.0$, what is the equilibrium constant expression and value for the overall reaction $A(g) \rightleftharpoons C(g)$?
   Answer: $K_{overall} = \frac{[C]}{[A]} = 6.0$ When you encounter coupled equilibria, you're dealing with a sequence of reactions where the product of one becomes the reactant of the next. The key insight is that equilibrium constants multiply when reactions are added together. For the overall reaction $A(g) \rightleftharpoons C(g)$, you can think of it as adding the two given reactions: - $A(g) \rightleftharpoons B(g)$ with $K_1 = 2.0$ - $B(g) \rightleftharpoons C(g)$ with $K_2 = 3.0$ When you add these reactions, B cancels out as an intermediate, leaving $A(g) \rightleftharpoons C(g)$. The equilibrium constant for this overall reaction is $K_{overall} = K_1 \times K_2 = 2.0 \times 3.0 = 6.0$. The expression is $K_{overall} = \frac{[C]}{[A]} = 6.0$, making answer B correct. Answer A incorrectly adds the equilibrium constants (2.0 + 3.0 = 5.0) instead of multiplying them. Answer C gives 1.5, which would result from dividing $K_2$ by $K_1$ (3.0/2.0), a common error when students confuse the order of operations. Answer D has the wrong equilibrium expression entirely—equilibrium constants are ratios of concentrations, not products of all species involved. Remember this pattern: when reactions are coupled in sequence, multiply their equilibrium constants to find the overall constant. This applies whenever you're combining equilibria where intermediates cancel out.
10. For the gas-phase reaction $2A(g) \rightleftharpoons B(g) + 3C(g)$, the equilibrium constant $K_p = 0.85$ at 500 K. What would be the equilibrium constant expression and value for the reaction $B(g) + 3C(g) \rightleftharpoons 2A(g)$ at the same temperature?
    Answer: $K_p = \frac{P_A^2}{P_B \cdot P_C^3} = 1.18$ When you encounter equilibrium problems involving reverse reactions, remember that reversing a chemical equation creates a reciprocal relationship between the equilibrium constants. For the original reaction $2A(g) \rightleftharpoons B(g) + 3C(g)$, the equilibrium constant expression is $K_p = \frac{P_B \cdot P_C^3}{P_A^2} = 0.85$. When we reverse this reaction to get $B(g) + 3C(g) \rightleftharpoons 2A(g)$, the products and reactants switch places in the equilibrium expression. The new expression becomes $K_p = \frac{P_A^2}{P_B \cdot P_C^3}$, and the numerical value becomes the reciprocal of the original: $K_p = \frac{1}{0.85} = 1.18$. Choice A correctly shows both the proper expression and value. Choice B has the correct expression but incorrectly keeps the original K value of 0.85, missing that reverse reactions require reciprocal K values. Choice C maintains the original expression $\frac{P_B \cdot P_C^3}{P_A^2}$ without accounting for the reversed equation, though it correctly calculates the reciprocal value. Choice D combines both errors—keeping the original expression and the original K value. Study tip: Always remember the "flip and flip" rule for reverse reactions: flip the equilibrium expression (products become reactants and vice versa) and flip the K value (take its reciprocal). This relationship holds for any equilibrium constant, whether $K_c$, $K_p$, or others.
11. The equilibrium $Cu^{2+}(aq) + 4NH_3(aq) \rightleftharpoons Cu(NH_3)_4^{2+}(aq)$ has $K_f = 1.1 \times 10^{13}$. What is the equilibrium constant expression for the dissociation reaction $Cu(NH_3)_4^{2+}(aq) \rightleftharpoons Cu^{2+}(aq) + 4NH_3(aq)$?
    Answer: $K_d = \frac{[Cu^{2+}][NH_3]^4}{[Cu(NH_3)_4^{2+}]} = 9.1 \times 10^{-14}$ When you encounter equilibrium problems involving formation and dissociation reactions, remember that these are reverse processes with equilibrium constants that are mathematical reciprocals of each other. The formation reaction has $K_f = 1.1 \times 10^{13}$, which represents the equilibrium constant for creating the complex ion from its components. The dissociation reaction is simply the reverse process - breaking apart the complex ion back into its original components. For any equilibrium expression, you write the ratio with products in the numerator and reactants in the denominator, each raised to their stoichiometric coefficients. For the dissociation reaction $Cu(NH_3)_4^{2+}(aq) \rightleftharpoons Cu^{2+}(aq) + 4NH_3(aq)$, the expression is $K_d = \frac{[Cu^{2+}][NH_3]^4}{[Cu(NH_3)_4^{2+}]}$. Since this is the reverse of the formation reaction, $K_d = \frac{1}{K_f} = \frac{1}{1.1 \times 10^{13}} = 9.1 \times 10^{-14}$. Option A correctly shows both the proper equilibrium expression and calculated value. Option B has the expression inverted (it's actually the formation constant expression). Option C uses the correct expression but keeps the formation constant value instead of taking the reciprocal. Option D omits the denominator entirely, which violates the fundamental structure of equilibrium expressions. Study tip: When dealing with reverse reactions, always remember: same expression format (products over reactants), but take the reciprocal of the given equilibrium constant.
12. A student studies the equilibrium $CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)$ and writes $K_c = \frac{[CO_2][H_2]}{[CO][H_2O]}$. The student then claims that because water is involved, the equilibrium expression should be modified. Evaluate this claim.
    Answer: The claim is incorrect; gaseous water is included in equilibrium expressions like any other gas When you encounter equilibrium expressions, the key principle is that you include all gaseous and aqueous species using their concentrations or partial pressures, while omitting pure solids and pure liquids (which have constant activities). The student's original expression $K_c = \frac{[CO_2][H_2]}{[CO][H_2O]}$ is completely correct. Notice that water appears as $H_2O(g)$ in the balanced equation—it's in the gas phase, not liquid phase. Gaseous water behaves just like any other gas in equilibrium expressions, so you include its concentration exactly as written. The student's concern about modifying the expression because "water is involved" reflects a common misconception. Looking at the wrong answers: (A) incorrectly assumes water should be omitted, but this only applies to liquid water, not gaseous water. The phase matters critically here. (C) suggests replacing water with vapor pressure, but this confuses $K_c$ (concentration-based) with $K_p$ (pressure-based) expressions—you don't mix concentration and pressure terms in the same expression. (D) mentions activities, which is technically more rigorous than concentrations, but this doesn't address the student's specific concern about water, and using activities isn't necessary for this level of analysis. The correct answer is (B): gaseous water is included in equilibrium expressions just like any other gas. Study tip: Always check the phase of each species in the balanced equation. Only pure solids and pure liquids get omitted from equilibrium expressions—gases and aqueous species are always included, regardless of their chemical identity.
13. For the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, both $K_c$ and $K_p$ can be used to describe the equilibrium. At 400°C, which statement correctly describes the relationship between these constants?
    Answer: $K_p < K_c$ because $\Delta n < 0$ for this reaction When you encounter equilibrium constant relationships, you need to understand how $K_p$ and $K_c$ are connected through the ideal gas law. These constants are related by the equation: $K_p = K_c(RT)^{\Delta n}$, where $\Delta n$ is the change in moles of gas (products minus reactants). For this reaction, $\Delta n = 2 - (1 + 3) = -2$. Since $\Delta n$ is negative, the equation becomes $K_p = K_c(RT)^{-2} = \frac{K_c}{(RT)^2}$. At 400°C (673 K), $RT$ is a large positive number, so $(RT)^2$ is much greater than 1. This means $K_p < K_c$. Choice B is correct because when $\Delta n < 0$, the equilibrium constant $K_p$ will always be smaller than $K_c$ at any temperature above absolute zero. Choice A incorrectly suggests that temperature alone determines the relationship, ignoring the crucial role of $\Delta n$. Choice C makes the common mistake of assuming that because all species are gases, the constants are equal—this only happens when $\Delta n = 0$. Choice D incorrectly claims pressure effects dominate at high temperature, but the relationship depends on the stoichiometric change in gas molecules, not pressure effects. Remember this pattern: when the number of gas molecules decreases in a reaction ($\Delta n < 0$), $K_p < K_c$. When it increases ($\Delta n > 0$), $K_p > K_c$. Only when $\Delta n = 0$ do they equal each other.
14. A student writes the equilibrium expression for $2H_2S(g) + SO_2(g) \rightleftharpoons 3S(s) + 2H_2O(g)$ as $K = \frac{[S]^3[H_2O]^2}{[H_2S]^2[SO_2]}$. What correction should be made to this expression?
    Answer: Remove $[S]^3$ from the numerator because sulfur is a solid When writing equilibrium expressions, you must understand which species to include based on their physical states. The equilibrium constant K only includes concentrations of gases and aqueous solutions—pure solids and pure liquids are omitted because their concentrations remain essentially constant. The correct equilibrium expression should be $K = \frac{[H_2O]^2}{[H_2S]^2[SO_2]}$. Notice that the $[S]^3$ term is completely removed from the numerator because sulfur exists as a pure solid. The "concentration" of a pure solid doesn't change during the reaction—it's always the same regardless of how much solid is present. Therefore, this constant value is incorporated into the equilibrium constant itself rather than appearing explicitly in the expression. Answer A correctly identifies that $[S]^3$ must be removed because sulfur is a solid. Answer B is wrong because the exponents should indeed match the stoichiometric coefficients from the balanced equation—this is a fundamental rule for writing equilibrium expressions. Answer C is incorrect because you never include density or any other property of pure solids in equilibrium expressions. Answer D is wrong because while you could use partial pressures instead of concentrations for gases (giving you $K_p$ instead of $K_c$), this doesn't address the main error of including the solid sulfur. Remember this key rule: when writing equilibrium expressions, include only gases and aqueous species, and always use the stoichiometric coefficients as exponents. Pure solids and liquids stay out of the expression entirely.
15. For the equilibrium $2NO(g) + Br_2(g) \rightleftharpoons 2NOBr(g)$, a student measures concentrations at equilibrium and calculates $Q = \frac{[NOBr]^2}{[NO]^2[Br_2]}$. The student then compares Q to the known $K_c$ value. What information does this comparison provide?
    Answer: If $Q = K_c$, the system is at equilibrium and no net reaction occurs When you encounter equilibrium problems involving the reaction quotient Q and equilibrium constant $K_c$, you're dealing with a fundamental tool for predicting reaction direction and determining equilibrium status. The reaction quotient Q has the same mathematical form as the equilibrium constant but uses current concentrations rather than equilibrium concentrations. Comparing Q to $K_c$ reveals the system's status: when $Q = K_c$, the system is at equilibrium with no net change in concentrations over time. This makes choice A correct. Let's examine why the other options are flawed. Choice B incorrectly states that when $Q > K_c$, the forward reaction is favored. Actually, when Q exceeds $K_c$, there's too much product relative to reactants, so the reverse reaction is favored to reduce Q back to $K_c$. Choice C reverses the logic: when $Q < K_c$, there's too little product, so the forward reaction proceeds until Q increases to equal $K_c$. Choice D is simply incorrect—the ratio $Q/K_c$ doesn't give percentage completion; it only indicates direction and how far from equilibrium the system currently sits. Remember this key pattern: Q acts like a compass pointing toward equilibrium. If $Q < K_c$, the reaction shifts forward (right); if $Q > K_c$, it shifts backward (left); if $Q = K_c$, you've reached your destination—equilibrium. This comparison never tells you about reaction completion, only direction and equilibrium status.
16. Consider the equilibrium $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$ with $K_p = 2.3 \times 10^{-4}$ at 250°C. What is the relationship between $K_p$ and $K_c$ for this reaction?
    Answer: $K_p = K_c(RT)$ When you encounter equilibrium problems involving both $K_p$ and $K_c$, you need to understand how these equilibrium constants relate through the ideal gas law. The key relationship is $K_p = K_c(RT)^{\Delta n}$, where $\Delta n$ is the change in moles of gas from reactants to products. For this reaction, $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$, you have 1 mole of gaseous reactant and 2 moles of gaseous products. Therefore, $\Delta n = (1 + 1) - 1 = +1$. Substituting into the general relationship: $K_p = K_c(RT)^1 = K_c(RT)$, which matches answer choice B. Looking at the wrong answers: Choice A ($K_p = K_c$) would only be true when $\Delta n = 0$, meaning equal moles of gaseous reactants and products. That's not the case here. Choice C ($K_p = K_c(RT)^{-1}$) represents $\Delta n = -1$, which would occur if you had more gaseous reactants than products—the opposite of this reaction. Choice D ($K_p = K_c(RT)^2$) corresponds to $\Delta n = +2$, meaning you'd need two more moles of gaseous products than reactants. Remember this pattern: always count the moles of gaseous species on each side of the equation to find $\Delta n$, then apply $K_p = K_c(RT)^{\Delta n}$. The sign and magnitude of $\Delta n$ directly determine the relationship between $K_p$ and $K_c$.
17. For a general equilibrium $aA + bB \rightleftharpoons cC + dD$, a student writes $K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$. Under what conditions is this expression valid?
    Answer: Only when all species are in the same phase and activities equal concentrations When you encounter equilibrium constant expressions, remember that the standard form $K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$ is actually a simplified version of the thermodynamically rigorous expression, which uses activities rather than concentrations. The correct answer is C because this concentration-based expression is only valid when activities can be approximated by concentrations. This occurs when all species are in the same phase (avoiding complications from different activity coefficient relationships across phases) and when the activity coefficients are approximately 1, meaning activities equal concentrations. Let's examine why the other options are incorrect: A) This is too restrictive. The expression works for other situations beyond just ideal gases, such as dilute solutions where activity coefficients approach unity. B) This is also too narrow. While dilute aqueous solutions are one valid condition (since activity coefficients ≈ 1 in dilute solutions), the expression can work in other scenarios too, like ideal gas mixtures or dilute solutions in non-aqueous solvents. D) This is incorrect because at high concentrations, in mixed phases, or under non-ideal conditions, activity coefficients deviate significantly from 1. In these cases, you must use the rigorous form: $K = \frac{a_C^c \cdot a_D^d}{a_A^a \cdot a_B^b}$ where $a$ represents activity. Study tip: Remember that concentration-based equilibrium expressions are approximations that work best under "ideal" conditions—same phase and low concentrations. When conditions become non-ideal, always consider whether activities might differ significantly from concentrations.
18. The equilibrium $BaF_2(s) \rightleftharpoons Ba^{2+}(aq) + 2F^-(aq)$ has $K_{sp} = [Ba^{2+}][F^-]^2$. A student argues that this expression is incorrect because it doesn't include the solid $BaF_2$. Which statement best addresses this argument?
    Answer: The student is incorrect; pure solids are omitted from equilibrium expressions When you encounter equilibrium expressions, especially for solubility equilibria, remember that the expression only includes species whose concentrations can actually change during the reaction. The correct $K_{sp}$ expression omits solid $BaF_2$ because pure solids have constant "concentration" - they don't dissolve into the solution in a way that changes their effective concentration in the equilibrium expression. The concentration of a pure solid is essentially its density divided by molar mass, which remains constant as long as solid is present. Including it would make $K_{sp}$ dependent on an unchanging value, which defeats the purpose of an equilibrium constant. Looking at the incorrect choices: Choice A suggests including $[BaF_2]$ in the denominator, but this would create a meaningless expression since the "concentration" of pure solid is constant and doesn't reflect the equilibrium position. Choice C mentions surface area, but while surface area affects the rate of dissolution, it doesn't appear in the equilibrium expression - $K_{sp}$ only depends on temperature. Choice D incorrectly states that only cation concentrations matter, but $K_{sp}$ expressions include all dissolved ions, both cations and anions, raised to appropriate powers based on stoichiometry. The answer is B - pure solids (and pure liquids) are always omitted from equilibrium expressions because their "concentrations" remain constant. Study tip: Remember the rule: only include gases and aqueous species in equilibrium expressions. Pure solids and liquids are always omitted, regardless of the type of equilibrium reaction.
19. For the reaction $A(g) + 2B(g) \rightleftharpoons C(g)$, a student writes $K_p = \frac{P_C}{P_A \cdot P_B^2}$. The student then claims that $K_c = \frac{[C]}{[A][B]^2}$ and that $K_p = K_c$ for this reaction. Evaluate these claims.
    Answer: Both expressions are correct, but $K_p ≠ K_c$ because $\Delta n ≠ 0$ When you encounter equilibrium problems involving both $K_p$ and $K_c$, you need to evaluate the expressions and understand the relationship between these two constants. Let's examine the student's expressions. For $K_p = \frac{P_C}{P_A \cdot P_B^2}$, this correctly follows the rule that equilibrium expressions use products raised to their stoichiometric coefficients divided by reactants raised to their coefficients. Similarly, $K_c = \frac{[C]}{[A][B]^2}$ properly applies the same stoichiometric coefficients to molar concentrations. Both expressions are indeed correct. However, the relationship between $K_p$ and $K_c$ depends on the change in moles of gas ($\Delta n$). The relationship is $K_p = K_c(RT)^{\Delta n}$, where $\Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants}$. For this reaction, $\Delta n = 1 - (1 + 2) = -2$. Since $\Delta n \neq 0$, we have $K_p = K_c(RT)^{-2}$, so $K_p \neq K_c$. Choice A incorrectly claims $K_p = K_c$ despite both expressions being right. Choice C wrongly suggests the $K_c$ expression lacks the proper exponent on [B]. Choice D incorrectly implies the $K_p$ expression needs atmospheric pressure corrections when it's already properly written. Choice B correctly identifies that both expressions are right but $K_p \neq K_c$ because $\Delta n \neq 0$. Remember: $K_p = K_c$ only when $\Delta n = 0$ (equal moles of gaseous reactants and products). Always calculate $\Delta n$ to determine their relationship.
20. The equilibrium $Ag^+(aq) + 2NH_3(aq) \rightleftharpoons Ag(NH_3)_2^+(aq)$ has a formation constant $K_f = 1.7 \times 10^7$. Which expression correctly represents this equilibrium constant?
    Answer: $K_f = \frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}$ When you encounter equilibrium expressions involving complex ion formation, you're working with formation constants (Kf), which follow the same rules as any equilibrium constant: products over reactants, with each concentration raised to its stoichiometric coefficient. For the equilibrium $Ag^+(aq) + 2NH_3(aq) \rightleftharpoons Ag(NH_3)_2^+(aq)$, the formation constant expression places the product (the complex ion) in the numerator and the reactants in the denominator. Since two moles of NH₃ participate in the reaction, its concentration gets squared. This gives us $K_f = \frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}$, which is answer choice B. Let's examine why the other options are incorrect. Choice A has the expression flipped upside down—this would actually represent the dissociation constant (Kd), which is the reciprocal of the formation constant. Choice C omits the exponent on [NH₃], ignoring the stoichiometric coefficient of 2 in the balanced equation. Choice D multiplies all concentrations together rather than creating a ratio, which doesn't follow the equilibrium constant format at all. Here's your key strategy: For any equilibrium constant expression, always write products over reactants, and don't forget the stoichiometric coefficients become exponents. Formation constants specifically describe the formation of complex ions from simpler components, so the complex ion always goes in the numerator. When in doubt, remember that Kf and Kd are reciprocals of each other.

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