Equilibrium Constants (College Board AP® Chemistry): Flashcards
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Define reaction quotient (Q).
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Define reaction quotient (Q).
The ratio of the concentrations (Qc) or partial pressures (Qp) of products to reactants at any point during a reaction, calculated using the same expression as the equilibrium constant K.
How does Q differ from K, and why is comparing them useful?
Q is calculated at any point in the reaction using current concentrations or partial pressures, whereas K is calculated only at equilibrium. Comparing Q to K predicts the direction of net reaction: if Q < K the forward reaction is favored; if Q > K the reverse reaction is favored; if Q = K the system is at equilibrium.
True or False?
Solids and pure liquids are included in the expression for Qc.
False.
Solids and pure liquids are omitted from Q and K expressions because their concentrations do not change with the amount of substance present.
If Q < K for a reaction at a given moment, in which direction will the reaction proceed, and why?
The reaction will proceed in the forward direction (toward products). Because Q is smaller than K, the product-to-reactant ratio is lower than at equilibrium, so the system shifts to produce more products until Q equals K.
True or False?
Both Q and K are unitless values.
True.
Although concentrations (M) or partial pressures (atm) are substituted into Q and K expressions, the resulting values are unitless.
For the reaction aA + bB ⇌ cC + dD, the expression for Qc is Qc = .......... / .......... , where brackets denote molar concentration.
For the reaction aA + bB ⇌ cC + dD, the expression for Qc is Qc = [C]c[D]d / [A]a[B]b , where brackets denote molar concentration.
Define Kc (the equilibrium constant in terms of concentration).
For the reaction aA + bB ⇌ cC + dD, Kc = [C]c[D]d / [A]a[B]b, where square brackets denote molar concentration (M). It is a unitless value that is constant at a given temperature.
What is the relationship between Kp and Kc, and what does each term represent?
Kp = Kc(RT)Δn\n\nWhere R = 0.08206 L·atm/mol·K, T = temperature in K and Δn = moles of gaseous products − moles of gaseous reactants. Kp uses partial pressures (atm) and Kc uses molar concentrations (M). Kp and Kc are equal when Δn = 0.
True or False?
The equilibrium constant Kc has units of mol/L.
False.
Kc is a unitless value. Although concentrations in M are substituted into the expression, the resulting equilibrium constant carries no units.
Define Kp (the equilibrium constant in terms of partial pressure).
For the reaction aA + bB ⇌ cC + dD, Kp = (Pc)c(PD)d / (PA)a(PB)b, where P denotes partial pressure in atm. It is a unitless value applicable to gas-phase equilibria.
True or False?
Brackets should be used in a Kp expression to denote partial pressures.
False.
Brackets are used only in Kc expressions to denote molar concentration. Kp expressions use partial pressures without brackets.
For the reaction 2NO2 (g) ⇌ 2NO (g) + O2 (g), what information is needed to calculate Kp, and how is it found?
The partial pressure of each gas at equilibrium is needed, found by multiplying each species' mole fraction by the total pressure. These values are substituted into Kp = (PNO)2(PO2) / (PNO2)2.
For the generic reaction aA + bB ⇌ cC + dD, the equilibrium expression for Kc is Kc = .......... / .......... , where square brackets represent molar concentration.
For the generic reaction aA + bB ⇌ cC + dD, the equilibrium expression for Kc is Kc = [C]c[D]d / [A]a[B]b , where square brackets represent molar concentration.
What does it mean when K >> 1 for a reaction?
Products are strongly favored at equilibrium. A very large K indicates that the forward reaction essentially proceeds to completion, with very little reactant remaining.
What does a very small equilibrium constant (K << 1) tell us about the extent of the forward reaction, and why?
A very small K means the forward reaction barely proceeds and reactants are strongly favored at equilibrium. Because K = [products]/[reactants], a value much less than 1 means the denominator (reactant concentrations) greatly exceeds the numerator (product concentrations) at equilibrium.
True or False?
When K > 1 for a reaction, products are favored over reactants at equilibrium.
True.
K > 1 means the numerator (product concentrations or partial pressures) is greater than the denominator (reactant concentrations or partial pressures) at equilibrium, so products are favored.
Two reactions at the same temperature have K values of 1.1 × 1033 and 4.6 × 1017. Starting from equal initial concentrations, which reaction produces a greater proportion of products at equilibrium, and why?
The reaction with K = 1.1 × 1033 produces a greater proportion of products. The larger the K value at the same temperature, the more the equilibrium lies toward products, so a higher fraction of reactants is converted at equilibrium.
True or False?
A large equilibrium constant (K >> 1) means the reaction reaches equilibrium quickly.
False.
The magnitude of K describes the position of equilibrium (how far the reaction proceeds), not how fast it reaches equilibrium. Reaction rate is determined by kinetic factors such as activation energy, not by K.
When K < 1 for a reaction, .......... are favored at equilibrium, and when K > 1, .......... are favored at equilibrium.
When K < 1 for a reaction, reactants are favored at equilibrium, and when K > 1, products are favored at equilibrium.
What is the equilibrium constant for a reversed reaction in terms of the original K?
The equilibrium constant for a reversed reaction is the reciprocal of the original: Knew = 1 / Koriginal. Reversing the reaction swaps products and reactants in the K expression.
If the stoichiometric coefficients of a reaction are multiplied by a factor c, how does this affect the equilibrium constant K, and why is the effect different from what happens to ΔH°?
Multiplying all stoichiometric coefficients by c raises K to the power of c: Knew = (Koriginal)^c. This contrasts with ΔH°, which scales linearly by c, because K is a product of concentration terms each raised to stoichiometric powers, so scaling coefficients exponentiates the entire expression.
True or False?
If a reaction is reversed, the new equilibrium constant is double the original K.
False.
Reversing a reaction gives Knew = 1 / Koriginal (the reciprocal), not double. Doubling the stoichiometric coefficients would give Knew = (Koriginal)2.
How is the equilibrium constant for a combined (Hess's Law-style) reaction determined from the K values of the individual steps?
The equilibrium constant for the overall reaction is the product of the K values for each individual step: K_overall = K1 × K2 × ... This follows because combining equations adds their free energies, which corresponds to multiplying their equilibrium constants.
True or False?
When the stoichiometric coefficients of a reaction are multiplied by 3, the equilibrium constant is also multiplied by 3.
False.
The new K is the original K raised to the power of 3: Knew = (Koriginal)3. Multiplying by 3 would apply to ΔH°, not to K.
What is the equilibrium constant for a reaction whose stoichiometric coefficients have been scaled by a factor c?
Knew = (Koriginal)^c. Scaling the stoichiometric coefficients by c raises the equilibrium constant to the power c, because each concentration or pressure term in the K expression is raised to the stoichiometric coefficient.
When two chemical equations are combined, the equilibrium constant for the overall reaction is the .......... of the K values of each individual step.
When two chemical equations are combined, the equilibrium constant for the overall reaction is the product of the K values of each individual step.
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