Properties of the Equilibrium Constant - College Chemistry

Consider the equilibrium reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ with $K_c = 0.60$ at 500°C. If the equation is multiplied by 2 to give $2N_2(g) + 6H_2(g) \rightleftharpoons 4NH_3(g)$ , what is the new equilibrium constant?
Answer: 0.36 When you encounter questions about manipulating chemical equilibrium expressions, remember that the equilibrium constant changes in predictable ways based on how you modify the balanced equation. For the original reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ , the equilibrium constant is $K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = 0.60$ . When you multiply an entire equation by a coefficient, you raise the equilibrium constant to that power. Since we're multiplying by 2, the new equilibrium constant becomes $K_c^2 = (0.60)^2 = 0.36$ . This happens because the new equilibrium expression is $K_{c,new} = \frac{[NH_3]^4}{[N_2]^2[H_2]^6}$ , which equals $\left(\frac{[NH_3]^2}{[N_2][H_2]^3}\right)^2 = K_c^2$ . Option A (0.30) represents the error of dividing the original constant by 2 instead of squaring it. Option C (0.60) incorrectly assumes the equilibrium constant doesn't change when you modify the equation. Option D (1.2) comes from multiplying by 2 instead of raising to the power of 2. The correct answer is B (0.36). Remember this key rule: when you multiply a balanced equation by a factor n, raise the equilibrium constant to the *n*th power. When you reverse an equation, take the reciprocal of K. These transformations are essential for solving equilibrium problems involving multiple reaction steps.
The equilibrium constant for reaction (1) $A \rightleftharpoons B$ is $K_1 = 2.0$ , and for reaction (2) $B \rightleftharpoons C$ is $K_2 = 0.50$ . What is the equilibrium constant for the overall reaction $A \rightleftharpoons C$ ?
Answer: 1.0 When you encounter sequential equilibrium reactions, you need to understand how equilibrium constants combine when reactions are coupled together. For sequential reactions where the product of one reaction becomes the reactant of the next, you multiply the equilibrium constants. Here's why: when you add the two given reactions, $A \rightleftharpoons B$ (with $K_1 = 2.0$ ) and $B \rightleftharpoons C$ (with $K_2 = 0.50$ ), the intermediate species B cancels out, giving you the overall reaction $A \rightleftharpoons C$ . The equilibrium constant for the overall reaction is $K_{overall} = K_1 \times K_2 = 2.0 \times 0.50 = 1.0$ . This works because equilibrium expressions are ratios of concentrations, and when you multiply them, the intermediate B terms algebraically cancel out. Looking at the incorrect choices: (A) 0.25 results from dividing $K_2$ by $K_1$ instead of multiplying—this reverses the logic. (C) 1.5 comes from averaging the two constants, which has no basis in equilibrium theory. (D) 2.5 results from adding the constants, but equilibrium constants multiply, not add, when reactions are combined sequentially. The correct answer is B) 1.0. Study tip: Remember the rule "multiply for sequential, raise to powers for coefficients." When reactions occur in sequence (product of first becomes reactant of second), always multiply their equilibrium constants. This is one of the most tested concepts in equilibrium chemistry.
Which statement about the magnitude of equilibrium constants is CORRECT?
Answer: A large $K_c$ value ( $K_c >> 1$ ) indicates that products are strongly favored at equilibrium When you encounter questions about equilibrium constants, focus on what $K_c$ tells you about the relative concentrations of products versus reactants at equilibrium, not reaction speed or completion. The equilibrium constant expression $K_c = \frac{[products]}{[reactants]}$ reveals the position of equilibrium. When $K_c >> 1$ , the numerator (product concentrations) is much larger than the denominator (reactant concentrations), meaning products are heavily favored. This makes option A correct. Let's examine why the other choices are wrong: Option B misunderstands what $K_c << 1$ means. When $K_c$ is very small, reactants are favored over products at equilibrium - the reaction doesn't go to completion, it barely proceeds forward at all. Option C confuses concentration with mass. When $K_c = 1$ , the concentrations of products and reactants are equal (after accounting for stoichiometry), but this doesn't mean equal masses. Different compounds have different molar masses, so equal concentrations don't translate to equal masses. Option D is a classic trap that confuses thermodynamics with kinetics. The magnitude of $K_c$ tells you nothing about reaction rate or how quickly equilibrium is reached - that's determined by activation energy and reaction mechanisms, not the equilibrium position. Remember this key distinction: $K_c$ describes "how far" a reaction proceeds (thermodynamics), while reaction rates describe "how fast" it gets there (kinetics). These are completely independent concepts in chemistry.
For the reaction $2A(g) + B(g) \rightleftharpoons 3C(g)$ , the equilibrium constant $K_c = 0.25$ at 298 K. If this reaction is reversed to give $3C(g) \rightleftharpoons 2A(g) + B(g)$ , what is the equilibrium constant for the reversed reaction?
Answer: 4.0 When you encounter a question about reversing chemical reactions, you're working with a fundamental relationship in equilibrium chemistry: the equilibrium constant for a reversed reaction is the reciprocal of the original equilibrium constant. For the original reaction $2A(g) + B(g) \rightleftharpoons 3C(g)$ , the equilibrium constant is: $K_c = \frac{[C]^3}{[A]^2[B]} = 0.25$ When you reverse this reaction to $3C(g) \rightleftharpoons 2A(g) + B(g)$ , the equilibrium expression becomes: $K_{c(reversed)} = \frac{[A]^2[B]}{[C]^3}$ Notice this is exactly the reciprocal of the original expression. Therefore: $K_{c(reversed)} = \frac{1}{K_c} = \frac{1}{0.25} = 4.0$ This makes D) 4.0 correct. A) 0.25 represents the misconception that the equilibrium constant stays the same when a reaction is reversed—it doesn't. B) 0.50 might result from incorrectly halving the original constant, which has no theoretical basis. C) 2.0 could come from mistakenly thinking you multiply by 2 instead of taking the reciprocal, or from calculation errors. Remember this key relationship: when you reverse a reaction, take the reciprocal of the equilibrium constant ( $K_{reversed} = \frac{1}{K_{original}}$ ). This pattern appears frequently in equilibrium problems, so master this reciprocal relationship—it's one of the most reliable rules in chemical equilibrium.
For the equilibrium $Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons FeSCN^{2+}(aq)$ , $K_c = 890$ at 25°C. If equal volumes of 0.20 M $Fe(NO_3)_3$ and 0.10 M $KSCN$ are mixed, what is the value of the reaction quotient $Q_c$ immediately after mixing?
Answer: 0 When you encounter equilibrium problems involving reaction quotients, remember that $Q_c$ tells you the current state of the system before it reaches equilibrium, while $K_c$ describes the equilibrium state. The reaction quotient has the same mathematical form as the equilibrium constant: $Q_c = \frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^-]}$ . The key insight is determining what concentrations to use immediately after mixing. When equal volumes of 0.20 M $Fe(NO_3)_3$ and 0.10 M $KSCN$ are mixed, each solution is diluted by half. This gives initial concentrations of 0.10 M $Fe^{3+}$ and 0.05 M $SCN^-$ . Critically, no $FeSCN^{2+}$ product exists yet—the reaction hasn't had time to proceed. Since $[FeSCN^{2+}] = 0$ immediately after mixing, $Q_c = \frac{0}{(0.10)(0.05)} = 0$ . Answer A is correct. Answer B (89) might tempt you if you incorrectly calculated the diluted concentrations or confused the setup. Answer C (445) could result from using undiluted concentrations in some flawed calculation—this is exactly half of $K_c$ , suggesting a dilution error. Answer D (890) is simply the equilibrium constant $K_c$ , but $Q_c$ only equals $K_c$ when the system is at equilibrium, not immediately after mixing. Remember: $Q_c = 0$ whenever you mix reactants with no product initially present. Compare $Q_c$ to $K_c$ to predict which direction the reaction will proceed.
At 25°C, the equilibrium constant $K_c = 1.8 \times 10^{-5}$ for the reaction $NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$ . What can be concluded about the position of equilibrium?
Answer: The equilibrium lies far to the left since $K_c < 1$ When you encounter equilibrium constant problems, the key insight is that the magnitude of $K_c$ directly tells you which side of the reaction is favored. The equilibrium constant compares the concentration of products to reactants at equilibrium. For this ammonia hydrolysis reaction, $K_c = 1.8 \times 10^{-5} = 0.000018$ , which is much smaller than 1. This small value means that at equilibrium, the concentration of products ( $NH_4^+$ and $OH^-$ ) is much lower than the concentration of reactants ( $NH_3$ ). Therefore, the equilibrium lies far to the left, favoring the reactants. Looking at the wrong answers: Choice A incorrectly states that $K_c > 1$ , but $1.8 \times 10^{-5}$ is clearly much less than 1. Choice C suggests the equilibrium is balanced because $K_c$ is "close to 1," but $0.000018$ is actually five orders of magnitude smaller than 1 – that's nowhere near close. Choice D claims you need concentration data, but that misses the point entirely: $K_c$ itself already incorporates equilibrium concentrations and tells you everything you need to know about the equilibrium position. Remember this pattern: $K_c >> 1$ means products dominate (equilibrium right), $K_c << 1$ means reactants dominate (equilibrium left), and $K_c ≈ 1$ means roughly equal amounts. Any $K_c$ value with negative exponents (like $10^{-5}$ ) immediately signals that reactants are heavily favored.
At 500 K, the equilibrium constant $K_c = 170$ for the reaction $CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)$ . If the temperature is increased to 600 K and $K_c$ becomes 54, what can be concluded about the reaction?
Answer: The reaction is exothermic because $K_c$ decreases with temperature When you encounter a question about how equilibrium constants change with temperature, you're dealing with Le Chatelier's principle and the relationship between thermodynamics and equilibrium position. The key insight is understanding what happens to $K_c$ when temperature changes for endothermic versus exothermic reactions. Here, $K_c$ decreases from 170 at 500 K to 54 at 600 K as temperature increases. For exothermic reactions, increasing temperature shifts the equilibrium toward reactants (away from products), which decreases $K_c$ . This matches exactly what we observe. Think of it this way: in an exothermic reaction, heat is essentially a "product." When you add more heat (increase temperature), Le Chatelier's principle says the equilibrium shifts left to consume that excess heat, reducing the ratio of products to reactants (which is what $K_c$ measures). Looking at the wrong answers: A incorrectly states that $K_c$ increases with temperature, which contradicts the given data. C correctly identifies that $K_c$ decreases but wrongly concludes the reaction is endothermic—for endothermic reactions, $K_c$ actually increases with temperature because heat favors the forward reaction. D makes both errors: claiming $K_c$ increases (it doesn't) and that this indicates an exothermic reaction. The correct answer is B: the reaction is exothermic because $K_c$ decreases with temperature. Study tip: Remember the pattern: $K_c$ increases with temperature for endothermic reactions, decreases for exothermic reactions. The direction of $K_c$ change tells you about the reaction's enthalpy.
For the gas-phase reaction $2A \rightleftharpoons B + C$ , the equilibrium constant $K_p = 0.85$ at 600 K. If the equilibrium partial pressures are $P_B = P_C = 1.2 \, atm$ , what is the partial pressure of $A$ at equilibrium?
Answer: 1.3 atm When you encounter gas-phase equilibrium problems, you need to set up the equilibrium expression using partial pressures and solve for the unknown quantity. For the reaction $2A \rightleftharpoons B + C$ , the equilibrium expression is: $K_p = \frac{P_B \times P_C}{P_A^2}$ Notice that $P_A$ is squared because the stoichiometric coefficient of A is 2. Given that $K_p = 0.85$ and $P_B = P_C = 1.2 \, \text{atm}$ , you can substitute and solve: $0.85 = \frac{(1.2)(1.2)}{P_A^2} = \frac{1.44}{P_A^2}$ Rearranging: $P_A^2 = \frac{1.44}{0.85} = 1.694$ Therefore: $P_A = \sqrt{1.694} = 1.3 \, \text{atm}$ This confirms answer choice C is correct. Looking at the wrong answers: A (0.59 atm) results from incorrectly taking the square root of 0.85/2.44 instead of 1.44/0.85. B (1.2 atm) assumes all partial pressures are equal, ignoring the stoichiometry. D (1.7 atm) comes from failing to take the square root of 1.694. Remember that stoichiometric coefficients become exponents in equilibrium expressions. Always double-check that you've applied the correct powers and performed the algebra carefully. Gas-phase equilibrium problems frequently test whether you properly account for stoichiometry in your $K_p$ expression.
The equilibrium constant $K_c = 4.2 \times 10^{-3}$ for the reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ at 298 K. If the initial concentration of $N_2O_4$ is 0.50 M and no $NO_2$ is initially present, which statement is correct about the equilibrium concentrations?
Answer: $[NO_2] < [N_2O_4]$ at equilibrium because $K_c < 1$ indicates reactants are favored When you encounter equilibrium problems with given $K_c$ values, the magnitude of the equilibrium constant tells you which species will predominate at equilibrium. Since $K_c = 4.2 \times 10^{-3} < 1$ , the equilibrium lies to the left, favoring reactants over products. To verify this quantitatively, set up an ICE table. Starting with 0.50 M $N_2O_4$ and 0 M $NO_2$ , let $x$ be the amount of $N_2O_4$ that dissociates. At equilibrium: $[N_2O_4] = 0.50 - x$ and $[NO_2] = 2x$ . Substituting into the equilibrium expression: $K_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(2x)^2}{0.50 - x} = 4.2 \times 10^{-3}$ Solving this equation yields $x \approx 0.023$ M, giving equilibrium concentrations of approximately $[N_2O_4] = 0.48$ M and $[NO_2] = 0.046$ M. Clearly, $[NO_2] < [N_2O_4]$ , confirming answer B. Choice A is wrong because reactions with small $K_c$ values don't proceed to completion—they reach equilibrium with mostly unreacted starting material. Choice C incorrectly assumes stoichiometry determines concentration ratios at equilibrium; stoichiometry only affects the rate at which concentrations change. Choice D makes the error of thinking that producing more moles of product automatically means higher product concentration at equilibrium. Remember: $K_c < 1$ means reactants dominate at equilibrium, while $K_c > 1$ means products dominate. The actual numerical value tells you how strongly the equilibrium favors one side.
The equilibrium constant for the reaction $A(g) \rightleftharpoons B(g) + C(g)$ is $K_1 = 0.50$ at 298 K. The equilibrium constant for the reaction $2B(g) \rightleftharpoons D(g)$ is $K_2 = 4.0$ at the same temperature. What is the equilibrium constant for the overall reaction $2A(g) \rightleftharpoons 2C(g) + D(g)$ ?
Answer: 1.0 When you encounter problems involving multiple equilibrium reactions that combine to form an overall reaction, you need to understand how equilibrium constants relate when reactions are added, multiplied, or reversed. To find the equilibrium constant for the target reaction $2A(g) \rightleftharpoons 2C(g) + D(g)$ , you must combine the given reactions algebraically. Start by manipulating the given equations: From reaction 1: $A(g) \rightleftharpoons B(g) + C(g)$ , $K_1 = 0.50$ From reaction 2: $2B(g) \rightleftharpoons D(g)$ , $K_2 = 4.0$ To get the target reaction, multiply reaction 1 by 2: $2A(g) \rightleftharpoons 2B(g) + 2C(g)$ When you multiply a reaction by a coefficient, you raise the equilibrium constant to that power: $K_1^2 = (0.50)^2 = 0.25$ Now add this doubled reaction 1 to reaction 2: $2A(g) \rightleftharpoons 2B(g) + 2C(g)$ $2B(g) \rightleftharpoons D(g)$ The $2B(g)$ cancels out, giving: $2A(g) \rightleftharpoons 2C(g) + D(g)$ When adding reactions, you multiply their equilibrium constants: $K_{overall} = K_1^2 \times K_2 = 0.25 \times 4.0 = 1.0$ Answer A (1.0) is correct. Answer B (2.0) incorrectly adds the constants. Answer C (4.5) mistakenly adds $K_1 + K_2$ . Answer D (8.0) incorrectly uses $K_1 \times K_2 \times 2^2$ . Remember: when combining equilibrium reactions, multiply constants for added reactions and raise to powers for multiplied reactions.
For the gas-phase reaction $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ , $K_c = 2.8 \times 10^{2}$ at 1000 K. What is the value of $K_p$ for this reaction at the same temperature? ( $R = 0.0821 \, L \cdot atm \cdot mol^{-1} \cdot K^{-1}$ )
Answer: $3.4 \times 10^{0}$ When you encounter equilibrium constant problems involving both $K_c$ and $K_p$ , you need to understand the relationship between these two expressions. $K_c$ uses molar concentrations while $K_p$ uses partial pressures, and they're connected by the equation: $K_p = K_c(RT)^{\Delta n}$ , where $\Delta n$ is the change in moles of gas. First, calculate $\Delta n$ by finding the difference between moles of gaseous products and reactants: $\Delta n = (2 \text{ mol } SO_3) - (2 \text{ mol } SO_2 + 1 \text{ mol } O_2) = 2 - 3 = -1$ . Now substitute into the equation: $K_p = (2.8 \times 10^2)(0.0821 \times 1000)^{-1} = (2.8 \times 10^2)(82.1)^{-1} = \frac{280}{82.1} = 3.4$ This gives us $K_p = 3.4 \times 10^0$ , which is answer A. Looking at the wrong answers: B ( $2.8 \times 10^2$ ) assumes $K_p = K_c$ , which only occurs when $\Delta n = 0$ . C ( $1.2 \times 10^4$ ) results from incorrectly using $\Delta n = +1$ instead of $-1$ . D ( $2.3 \times 10^4$ ) appears to involve calculation errors or misapplication of the formula. Study tip: Always calculate $\Delta n$ carefully by counting gas molecules on each side of the equation. Remember that when $\Delta n$ is negative, $K_p$ will be smaller than $K_c$ , and when positive, $K_p$ will be larger than $K_c$ .
At equilibrium for the reaction $2NOBr(g) \rightleftharpoons 2NO(g) + Br_2(g)$ , the concentrations are $[NOBr] = 0.40 \, M$ , $[NO] = 0.20 \, M$ , and $[Br_2] = 0.10 \, M$ . What is the value of $K_c$ ?
Answer: 0.025 When you encounter an equilibrium expression problem, you need to write the equilibrium constant expression using the balanced chemical equation, then substitute the given concentrations. For the reaction $2NOBr(g) \rightleftharpoons 2NO(g) + Br_2(g)$ , the equilibrium constant expression is: $K_c = \frac{[NO]^2[Br_2]}{[NOBr]^2}$ Notice that the exponents match the stoichiometric coefficients from the balanced equation. Products go in the numerator, reactants in the denominator. Substituting the given equilibrium concentrations: $K_c = \frac{(0.20)^2(0.10)}{(0.40)^2} = \frac{(0.040)(0.10)}{0.16} = \frac{0.004}{0.16} = 0.025$ This confirms answer A is correct. Looking at the wrong answers: B (0.050) might result from forgetting to square the NO concentration or making an arithmetic error in the calculation. C (0.10) could come from incorrectly using just the Br₂ concentration or setting up the expression incorrectly. D (0.20) might result from using the NO concentration directly without proper calculation or inverting part of the expression. The key strategy here is to always write out the $K_c$ expression first based on the balanced equation, double-check that exponents match stoichiometric coefficients, then carefully substitute and calculate. Remember: products over reactants, and small $K_c$ values (like 0.025) indicate the equilibrium lies toward reactants, which makes sense since NOBr has the highest concentration.
At a certain temperature, $K_c = 1.5$ for the reaction $2NO(g) + Br_2(g) \rightleftharpoons 2NOBr(g)$ . If at equilibrium $[NO] = 0.80 \, M$ , $[Br_2] = 0.60 \, M$ , and $[NOBr] = 0.40 \, M$ , which statement is correct?
Answer: The reaction quotient $Q_c < K_c$ , so the reaction will shift right When you encounter equilibrium problems with given concentrations, you need to determine whether the system is actually at equilibrium by comparing the reaction quotient ( $Q_c$ ) to the equilibrium constant ( $K_c$ ). First, calculate $Q_c$ using the same expression as $K_c$ , but with the given concentrations: $Q_c = \frac{[NOBr]^2}{[NO]^2[Br_2]} = \frac{(0.40)^2}{(0.80)^2(0.60)} = \frac{0.16}{0.64 \times 0.60} = \frac{0.16}{0.384} = 0.42$ Since $Q_c = 0.42 < K_c = 1.5$ , the reaction will shift right (toward products) to reach equilibrium. This confirms answer C is correct. Now let's examine why the other answers are wrong: A) is incorrect because $Q_c \neq K_c$ , so these concentrations don't represent equilibrium. B) is wrong because $Q_c < K_c$ , not $Q_c > K_c$ . When $Q_c > K_c$ , the reaction would shift left, but that's not the case here. D) reflects a fundamental misunderstanding—equilibrium concentrations don't need to be equal to each other. They only need to satisfy the equilibrium constant expression. Remember this key strategy: Always calculate $Q_c$ first when given concentrations in an equilibrium problem. Compare it to $K_c$ to determine the direction of shift: if $Q_c < K_c$ , shift right; if $Q_c > K_c$ , shift left; if $Q_c = K_c$ , you're at equilibrium.
At 500 K, the equilibrium constant $K_c = 0.061$ for the decomposition $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$ . In a 2.0 L container, the equilibrium mixture contains 0.15 mol $PCl_3$ and 0.15 mol $Cl_2$ . How many moles of $PCl_5$ are present at equilibrium?
Answer: 0.18 mol When you encounter equilibrium constant problems, you're working with the relationship between concentrations (or pressures) of reactants and products at equilibrium. The key is setting up the $K_c$ expression correctly and using the given information systematically. For this reaction, $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = 0.061$ . First, convert the given moles to concentrations by dividing by the 2.0 L volume: $[PCl_3] = [Cl_2] = \frac{0.15 \text{ mol}}{2.0 \text{ L}} = 0.075 \text{ M}$ . Substituting into the equilibrium expression: $0.061 = \frac{(0.075)(0.075)}{[PCl_5]}$ . Solving for $[PCl_5]$ : $[PCl_5] = \frac{(0.075)^2}{0.061} = \frac{0.005625}{0.061} = 0.092 \text{ M}$ . Converting back to moles: $0.092 \text{ M} \times 2.0 \text{ L} = 0.18$ mol. Answer A (0.092 mol) represents the common error of forgetting to convert concentration back to moles—this is the concentration value, not the number of moles. Answer B (0.15 mol) might result from assuming equal molar amounts of all species at equilibrium, which ignores the equilibrium constant. Answer D (0.92 mol) could come from calculation errors or misapplying the equilibrium expression. Always remember the two-step conversion process in equilibrium problems: convert given moles to molarity for calculations, then convert your final concentration back to moles if that's what the question asks for. Double-check your units throughout.
At a certain temperature, $K_c = 4.0 \times 10^{-2}$ for the reaction $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$ . What is $K_c$ for the reaction $\frac{1}{2}H_2(g) + \frac{1}{2}I_2(g) \rightleftharpoons HI(g)$ ?
Answer: 5.0 When you encounter questions about equilibrium constants and reaction manipulation, remember that the equilibrium constant changes predictably when you alter the chemical equation. The key is understanding how reversing reactions and changing stoichiometric coefficients affect $K_c$ . The original reaction has $K_c = 4.0 \times 10^{-2}$ for $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$ . The target reaction is $\frac{1}{2}H_2(g) + \frac{1}{2}I_2(g) \rightleftharpoons HI(g)$ . Notice two changes: the reaction is reversed (products become reactants), and all coefficients are divided by 4 (since the original has 2 HI, and the new reaction has 1 HI). When you reverse a reaction, the new $K_c$ becomes $\frac{1}{K_{original}}$ . When you divide all coefficients by a factor, you raise $K_c$ to the power equal to that factor. Here, reversing gives $\frac{1}{4.0 \times 10^{-2}} = 25$ . Then dividing coefficients by 4 means taking the fourth root: $K_c = (25)^{1/4} = (5^2)^{1/4} = 5^{1/2} = \sqrt{25} = 5.0$ . Choice A (0.020) incorrectly keeps the original $K_c$ value. Choice B (0.20) appears to reverse the reaction but ignores the coefficient change. Choice C (2.0) might result from taking the square root of the reversed $K_c$ instead of the fourth root. Remember this pattern: reversed reaction means $K_{new} = \frac{1}{K_{original}}$ , and coefficient changes mean raising $K_c$ to the power of the scaling factor. Always identify both transformations separately, then apply them sequentially.
For a general reaction $aA + bB \rightleftharpoons cC + dD$ , which expression correctly represents the equilibrium constant $K_c$ ?
Answer: $K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$ When you encounter equilibrium constant expressions, remember that they follow a universal pattern: products over reactants, with each concentration raised to its stoichiometric coefficient as an exponent. For any equilibrium reaction, the equilibrium constant $K_c$ is constructed by placing the concentrations of products in the numerator and reactants in the denominator. Each concentration is raised to the power of its coefficient from the balanced equation. For the reaction $aA + bB \rightleftharpoons cC + dD$ , products C and D go on top, while reactants A and B go on bottom, giving us $K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$ . This is answer choice B. Choice A reverses this relationship, incorrectly placing reactants over products. This is a common mistake that would give you the expression for $\frac{1}{K_c}$ rather than $K_c$ itself. Choice C includes the correct products-over-reactants setup but omits the crucial exponents, ignoring the stoichiometric coefficients entirely. This fundamentally misrepresents how equilibrium responds to concentration changes. Choice D uses subtraction instead of division, which has no basis in equilibrium theory and wouldn't produce a meaningful constant value. The key study tip: Always remember "products over reactants, coefficients become exponents." This pattern applies to every equilibrium constant expression you'll encounter, whether it's $K_c$ , $K_p$ , or $K_{eq}$ . The stoichiometric coefficients are essential—they reflect how the equilibrium position responds to concentration changes according to Le Chatelier's principle.
Which of the following statements about equilibrium constants is FALSE?
Answer: The equilibrium constant expression includes concentration terms for pure solids and liquids When you encounter questions about equilibrium constants, focus on the fundamental rules that govern how these constants are defined and calculated. The key issue here involves what gets included in equilibrium constant expressions. For any equilibrium reaction, the expression $K = \frac{[products]}{[reactants]}$ only includes species whose concentrations can actually change during the reaction. Pure solids and pure liquids have constant concentrations that don't change as the reaction proceeds, so they're omitted from the equilibrium expression and assigned an activity of 1. This makes option C false – pure solids and liquids are specifically excluded from equilibrium constant expressions. Let's examine why the other options are true. Option A is correct because $K$ depends only on the ratio of concentrations at equilibrium, not on where you started. You can begin with any initial concentrations and still reach the same equilibrium constant value. Option B is accurate since temperature changes affect the forward and reverse reaction rates differently, shifting the equilibrium position and changing $K$ . Option D reflects a mathematical relationship: if $K_{forward} = \frac{[products]}{[reactants]}$ , then $K_{reverse} = \frac{[reactants]}{[products]} = \frac{1}{K_{forward}}$ . Remember this pattern: equilibrium expressions only include gases and aqueous solutions – never pure solids, pure liquids, or solvents. When writing $K$ expressions, mentally cross out any pure phases before setting up your fraction.
For the equilibrium $2NO_2(g) \rightleftharpoons N_2O_4(g)$ , $K_c = 170$ at 298 K. If the concentrations at a particular moment are $[NO_2] = 0.10 \, M$ and $[N_2O_4] = 0.50 \, M$ , what is the value of the reaction quotient $Q_c$ ?
Answer: 50 When you encounter equilibrium problems, it's crucial to distinguish between the equilibrium constant ( $K_c$ ) and the reaction quotient ( $Q_c$ ). While $K_c$ uses equilibrium concentrations, $Q_c$ uses any given concentrations to determine which direction the reaction will proceed. The reaction quotient has the same mathematical form as the equilibrium constant. For the reaction $2NO_2(g) \rightleftharpoons N_2O_4(g)$ , the expression is: $Q_c = \frac{[N_2O_4]}{[NO_2]^2}$ Notice that $NO_2$ is squared because its coefficient is 2. Substituting the given concentrations: $Q_c = \frac{0.50}{(0.10)^2} = \frac{0.50}{0.01} = 50$ Looking at the wrong answers: Choice A (0.020) represents the reciprocal of the correct answer, suggesting you inverted the fraction. Choice B (0.50) is simply the concentration of $N_2O_4$ , indicating you forgot to include the denominator entirely. Choice C (5.0) results from forgetting to square the $NO_2$ concentration, calculating $\frac{0.50}{0.10}$ instead. The correct answer is D (50). Study tip: Always write out the $Q_c$ expression first, paying careful attention to stoichiometric coefficients becoming exponents. Remember that $Q_c < K_c$ means the reaction proceeds forward, $Q_c > K_c$ means it proceeds backward, and $Q_c = K_c$ indicates equilibrium. In this case, since $Q_c = 50$ is much less than $K_c = 170$ , the forward reaction is favored.
The equilibrium constant for the reaction $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$ is $K_c = 4.2 \times 10^{-31}$ at 298 K. What does this extremely small equilibrium constant indicate about the reaction?
Answer: At equilibrium, virtually no $NO$ will be present compared to $N_2$ and $O_2$ When you encounter equilibrium constant problems, focus on what the magnitude of $K_c$ tells you about the position of equilibrium. The equilibrium constant expression for this reaction is $K_c = \frac{[NO]^2}{[N_2][O_2]}$ . An extremely small $K_c$ value like $4.2 \times 10^{-31}$ means the numerator (products) must be much, much smaller than the denominator (reactants) at equilibrium. Since $K_c$ involves $[NO]^2$ in the numerator, this tiny value indicates that virtually no nitric oxide will form compared to the amounts of nitrogen and oxygen present. The equilibrium lies far to the left, favoring reactants. Option A is incorrect because $K_c$ tells you nothing about reaction rate or how quickly equilibrium is reached—only about the equilibrium position. Option B makes an unsupported claim about thermodynamics; while the reaction may indeed be endothermic, you cannot determine this from $K_c$ alone, and the small constant doesn't mean the reaction "will not occur"—it just means very little product forms. Option D contradicts what the small $K_c$ tells you; if the reaction went to completion, $K_c$ would be extremely large, not small. Therefore, C correctly identifies that the equilibrium mixture will contain essentially only $N_2$ and $O_2$ with negligible $NO$ . Study tip: Remember the $K_c$ magnitude rule: $K_c >> 1$ favors products, $K_c << 1$ favors reactants. Values around $10^{-30}$ indicate virtually no product formation.
For which of the following reactions is $K_p = K_c$ ?
Answer: $H_2(g) + Br_2(g) \rightleftharpoons 2HBr(g)$ When you encounter equilibrium constant problems, you need to understand the relationship between $K_p$ (pressure-based) and $K_c$ (concentration-based). The key equation is $K_p = K_c(RT)^{\Delta n}$ , where $\Delta n$ is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants). For $K_p = K_c$ , we need $(RT)^{\Delta n} = 1$ , which only occurs when $\Delta n = 0$ . This means the number of gas molecules on both sides of the equation must be equal. Let's check each reaction: A) $H_2(g) + Br_2(g) \rightleftharpoons 2HBr(g)$ : Here, 2 moles of reactants produce 2 moles of products, so $\Delta n = 2 - 2 = 0$ . This makes $K_p = K_c$ . B) $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ : We have 4 moles of reactants and 2 moles of products, giving $\Delta n = 2 - 4 = -2$ . Since $\Delta n \neq 0$ , $K_p \neq K_c$ . C) $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ : This gives us 3 moles of reactants and 2 moles of products, so $\Delta n = 2 - 3 = -1$ . Again, $K_p \neq K_c$ . D) $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$ : Here we have 1 mole of reactant and 2 moles of products, making $\Delta n = 2 - 1 = 1$ , so $K_p \neq K_c$ . Study tip: Always count gas molecules on each side first. When they're equal, $K_p = K_c$ immediately—no calculations needed.

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