4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
The diagram represents the equilibrium mixture for the reaction W(g) ⇌ Z(g) in a sealed container at 298 K. Which claim about ΔG° for the reaction at 298 K is correctly justified by the diagram?
Answer and reasoning
AΔG° > 0, because K = 8/2, which is greater than 1 A student who thinks ΔG° and K rise and fall together picks this. K = 8/2 = 4 is correct, but ΔG° = −RT ln K, so K > 1 gives a negative ΔG°.
BΔG° < 0, because K = 8/2, which is greater than 1Correct At equilibrium the box holds 8 Z and 2 W in the same volume, so K = [Z]/[W] = 8/2 = 4. Because K > 1, the products are favored at equilibrium, and ΔG° = −RT ln K is negative.
CΔG° > 0, because K = 2/8, which is smaller than 1 A student who writes K as reactant over product picks this. For W ⇌ Z, K = [Z]/[W] = 8/2 = 4, which is greater than 1, so ΔG° is negative.
DΔG° = 0, because the two gases are at equilibrium A student who thinks ΔG° is zero whenever a system is at equilibrium picks this. ΔG° is found from the value of K: ΔG° = −RT ln 4, which is negative. ΔG° would be zero only if K were 1, with equal amounts of W and Z.
A student wants to calculate the equilibrium constant, K, at 350 K for the reaction X(g) ⇌ Y(g), using the gas constant, R. Which single additional quantity is sufficient for the calculation?
Answer and reasoning
AΔG° for the reaction at 350 KCorrect K = e−ΔG°/RT. With R and T = 350 K known, the only other quantity in the equation is ΔG° at that temperature.
BΔH° for the reaction at 350 K A student who thinks the energy released decides whether products are favored picks this. ΔH° is only one contribution to ΔG° = ΔH° − TΔS°; without ΔS°, K cannot be calculated.
CΔS° for the reaction at 350 K A student who thinks an entropy increase of the reacting system decides whether products are favored picks this. ΔS° is only one contribution to ΔG° = ΔH° − TΔS°; without ΔH°, K cannot be calculated.
DEa for the reaction at 350 K A student who thinks a fast reaction has a large equilibrium constant picks this. The activation energy, Ea, affects how quickly equilibrium is reached; K depends on ΔG°, the difference between products and reactants.
Working No calculation. The relationship is K = e−ΔG°/RT. R is known and T = 350 K is given, so the one missing quantity is ΔG° at 350 K. ΔH° alone or ΔS° alone is not enough, because ΔG° = ΔH° − TΔS° needs both.
For a certain reaction, ΔG° = −4.0 kJ/molrxn at 298 K. At 298 K, RT = 2.5 kJ/mol. Which statement about the value of the equilibrium constant, K, at 298 K is correct?
Answer and reasoning
AK is smaller than 1 A student who matches a negative ΔG° with a K less than 1 picks this, evaluating e−1.6 = 0.20. The exponent is −ΔG°/RT = +1.6, so K = e1.6 ≈ 5.
BK is between 1 and 10Correct K = e−ΔG°/RT = e4.0/2.5 = e1.6. Because e¹ ≈ 2.7 and e² ≈ 7.4, K is about 5. ΔG° is negative and comparable in size with RT, so K is greater than 1 but not by a large factor.
CK is between 10 and 100 A student who treats ln and log as the same operation picks this, evaluating 101.6 ≈ 40. The inverse of the natural logarithm is e raised to the power: e1.6 ≈ 5.
DK is larger than 100 A student who thinks any reaction with ΔG° < 0 goes essentially to completion picks this. Here −ΔG°/RT is only 1.6, so K = e1.6 ≈ 5: the products are favored, but substantial reactant remains at equilibrium.
Working K = e−ΔG°/RT. −ΔG°/RT = (4.0 kJ/mol)/(2.5 kJ/mol) = 1.6. K = e1.6; since e¹ = 2.7 and e² = 7.4, K ≈ 5.0, which lies between 1 and 10.
For the reaction X(g) ⇌ Y(g), ΔG° = +5.4 kJ/molrxn at 298 K. Each numbered diagram shows ten particles in a sealed container. Which diagram best represents the equilibrium mixture at 298 K?
Answer and reasoning
ADiagram 1 A student who matches a positive ΔG° with a K greater than 1 picks the box with one X and nine Y. ΔG° > 0 gives K < 1 (here 0.11), so X, the reactant, is the major species.
BDiagram 2 A student who thinks a reaction with ΔG° > 0 does not occur picks the box with ten X and no Y. K = 0.11 is small but not zero: about one particle in ten is Y at equilibrium.
CDiagram 3Correct K = e−ΔG°/RT = e−5400/(8.314 × 298) = e−2.18 = 0.11, about 1/9. For X ⇌ Y, K = [Y]/[X], so the equilibrium mixture has about one Y for every nine X: reactants are favored, but some product is present.
DDiagram 4 A student who thinks reactant and product concentrations are equal at equilibrium picks the box with five X and five Y. That mixture corresponds to K = 1 and ΔG° = 0; here K = 0.11, so [Y]/[X] is about 1/9.
Working K = e−ΔG°/RT = e−5400 J/mol ÷ (8.314 J/(mol·K) × 298 K) = e−2.18 = 0.11 ≈ 1/9. K = [Y]/[X], so of ten particles about nine are X and one is Y.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
9.5.A.1 Thermodynamically favored and K Fix
Thermodynamically favored and K
A process is thermodynamically favored when ΔG° < 0, which means that the products are favored at equilibrium (K > 1) under standard conditions. A process with ΔG° > 0 favors the reactants at equilibrium (K < 1); some products still form, because K is less than 1 but not zero.
Students often think A system at equilibrium has ΔG° = 0, so equilibrium is possible only where ΔG° is zero, and a reaction with ΔG° different from zero is not at equilibrium. In fact No. ΔG° is a fixed value for a reaction at a given temperature, found from K by ΔG° = −RT ln K; it is zero only in the special case K = 1. A system can reach equilibrium whatever the value of ΔG°.
Students often think K is the amount of reactant divided by the amount of product at equilibrium. In fact No. For the reaction as written, K has the products in the numerator and the reactants in the denominator. An equilibrium mixture with more product than reactant particles in a one-to-one reaction has K > 1.
9.5.A.2 ΔG° = −RT ln K Fix
ΔG° = −RT ln K
The equation that relates the standard free energy change of a process to its equilibrium constant at the absolute temperature T. With R = 8.314 J/(mol·K) and T in kelvins, the equation gives ΔG° in joules per mole of reaction; ln is the natural logarithm.
K = e−ΔG°/RT
The same relationship solved for K. ΔG° and RT must be in the same energy unit (both in J/mol or both in kJ/mol) before the exponent −ΔG°/RT is evaluated, and T must be in kelvins.
Temperature in the ΔG°–K relationship
Standard conditions do not fix the temperature. ΔG° and K each have a value at every temperature, and the T used in ΔG° = −RT ln K is the temperature at which K was measured or is wanted, in kelvins.
Students often think ln and log are the same operation, so ΔG° = −RT ln K can be evaluated with the log key, and K can be found as 10−ΔG°/RT. In fact No. The equation uses the natural logarithm, ln, and its inverse is e raised to a power. Using log (base 10) gives a ΔG° that is too small in magnitude by a factor of 2.303, and using 10 raised to a power gives a K that is far too large or too small.
Students often think The temperature in ΔG° = −RT ln K can be entered in degrees Celsius, as it is given in the problem. In fact No. T is the absolute temperature in kelvins. Using a Celsius value changes the size of the answer and can even give T = 0 or a negative T.
9.5.A.3 Estimating K from ΔG° and RT Fix
Estimating K from ΔG° and RT
The size of ΔG° compared with RT (about 2.5 kJ/mol at 298 K) shows how far K is from 1. When ΔG° is near zero, K is close to 1 and the equilibrium mixture contains comparable amounts of reactants and products; when the magnitude of ΔG° is much larger than RT, K deviates strongly from 1.
ΔG° = 0 and K = 1
When ΔG° = 0 at some temperature, K = e⁰ = 1 at that temperature: neither reactants nor products are favored at equilibrium under standard conditions. A system can be at equilibrium at any temperature and for any value of ΔG°; equilibrium does not require ΔG° = 0.
Students often think K is proportional to −ΔG°, so K = 0 when ΔG° = 0 and no products are present at equilibrium. In fact No. When ΔG° = 0, K = e⁰ = 1. K is a ratio of concentrations or pressures and is positive for every value of ΔG°; it gets closer and closer to zero as ΔG° becomes very large and positive.
Students often think A reaction with ΔG° < 0 is favored, so it goes to completion and K is very large, whatever the size of ΔG°. In fact No. The extent of reaction depends on the size of ΔG° compared with RT. When ΔG° is negative but small compared with RT, K is only a little greater than 1 and the equilibrium mixture contains substantial amounts of both reactants and products.
9.5.A.4 Sign of ΔG° and position of equilibrium Fix
Sign of ΔG° and position of equilibrium
Processes with ΔG° < 0 favor products (K > 1), and processes with ΔG° > 0 favor reactants (K < 1). At a given temperature, the more negative ΔG° is, the larger K is.
Students often think A negative ΔG° goes with a small K: ΔG° < 0 means K < 1, and ΔG° > 0 means K > 1, so that ΔG° and K rise and fall together. In fact No. Because ΔG° = −RT ln K, a negative ΔG° requires ln K to be positive, so K > 1 and the products are favored at equilibrium. A positive ΔG° gives K < 1.
Students often think At equilibrium the concentrations of reactants and products are equal, whatever the values of K and ΔG°. In fact No. At equilibrium the forward and reverse rates are equal; the concentrations are constant and are fixed by K. They are equal only in special cases, such as K = 1 for a reaction of the type X ⇌ Y.
7 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 7
For the hypothetical reaction X(g) ⇌ Y(g), the equilibrium constant is K = 2.0 × 10⁻⁶ at 177°C. What is the value of ΔG° for the reaction at 177°C?
Answer and reasoning
A21 kJ/molrxn A student who treats ln and log as the same operation picks this: (8.314)(450)(5.70) = 2.1 × 10⁴ J/mol. The equation needs the natural logarithm, ln(2.0 × 10⁻⁶) = −13.12, which gives 49 kJ/molrxn.
B19 kJ/molrxn A student who enters the temperature in degrees Celsius, as given, picks this: (8.314)(177)(13.12) = 1.9 × 10⁴ J/mol. T must be in kelvins, 450 K: (8.314)(450)(13.12) = 4.9 × 10⁴ J/mol.
C33 kJ/molrxn A student who thinks the ° symbol fixes the temperature at 298 K picks this: (8.314)(298)(13.12) = 3.3 × 10⁴ J/mol. K was measured at 177°C, so T = 450 K and ΔG° = 49 kJ/molrxn.
D49 kJ/molrxnCorrect T = 177°C = 450 K. ΔG° = −RT ln K = −(8.314 J/(mol·K))(450 K) ln(2.0 × 10⁻⁶) = −(3741 J/mol)(−13.12) = +4.9 × 10⁴ J/mol = 49 kJ/molrxn. The value is positive, as expected for K < 1.
Working T = 177 + 273 = 450 K. ΔG° = −RT ln K. R = 8.314 J/(mol·K), ln(2.0 × 10⁻⁶) = −13.12. ΔG° = −(8.314 J/(mol·K))(450 K)(−13.12) = 4.91 × 10⁴ J/molrxn = 49 kJ/molrxn (2 significant figures).
For a certain reaction, ΔG° = −15 kJ/molrxn at 227°C. What is the value of the equilibrium constant, K, for the reaction at 227°C?
Answer and reasoning
A430 A student who thinks the ° symbol fixes the temperature at 298 K picks this: e15,000/(8.314 × 298) = e6.05. ΔG° was given at 227°C, so T = 500 K and K = 37.
B2800 A student who enters the temperature in degrees Celsius, as given, picks this: e15,000/(8.314 × 227) = e7.95. T must be in kelvins, 500 K, which gives K = 37.
C37Correct K = e−ΔG°/RT with ΔG° = −15,000 J/molrxn and T = 500 K: the exponent is +15,000/(8.314 × 500) = 3.61, so K = e3.61 = 37. K > 1, as expected for ΔG° < 0.
D0.027 A student who matches a negative ΔG° with a K less than 1 picks this, carrying the negative sign of ΔG° into the exponent: e−3.61. The exponent is −ΔG°/RT = +3.61, so K = 37.
Working T = 227 + 273 = 500 K. −ΔG°/RT = (15,000 J/mol)/(8.314 J/(mol·K) × 500 K) = 3.61. K = e3.61 = 37.
The graph shows how ΔG° for a reaction varies with temperature. Which statement about the equilibrium constant, K, of the reaction is supported by the graph?
Answer and reasoning
AAt 700 K, K > 1, so products are favored at equilibriumCorrect At 700 K the graph gives ΔG° = −40 kJ/molrxn. A process with ΔG° < 0 has K > 1, so the products are favored at equilibrium. (At 500 K, ΔG° = 0 and K = 1; at 300 K, ΔG° = +40 kJ/molrxn and K < 1.)
BAt 300 K, K > 1, so products are favored at equilibrium A student who thinks ΔG° and K rise and fall together picks this, matching the largest ΔG° on the graph with K > 1. At 300 K, ΔG° = +40 kJ/molrxn, and a positive ΔG° gives K < 1: the reactants are favored.
CAt 500 K, K = 0, so products are missing at equilibrium A student who thinks K is zero when ΔG° is zero picks this. At 500 K, ΔG° = 0, so K = e⁰ = 1: reactants and products are equally favored, and products are present.
DAt 500 K, ΔG° = 0, so only there does equilibrium occur A student who thinks ΔG° must be zero for a system to be at equilibrium picks this. The system can reach equilibrium at any temperature on the graph; ΔG° = 0 at 500 K means only that K = 1 at that temperature.
Working Read ΔG° from the line: +40 kJ/molrxn at 300 K, 0 at 500 K, −40 kJ/molrxn at 700 K. From ΔG° = −RT ln K: ΔG° > 0 gives ln K < 0, K < 1 (300 K); ΔG° = 0 gives K = 1 (500 K); ΔG° < 0 gives ln K > 0, K > 1 (700 K).
A student plans to determine ΔG° for the dissolution Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2 OH⁻(aq). The student will titrate a measured volume of filtered, saturated Ca(OH)₂ solution with standardized HCl(aq) to find [OH⁻], and from it Ksp. Which additional measurement must be part of the procedure so that ΔG° can be calculated?
Answer and reasoning
AThe mass of solid Ca(OH)₂ left undissolved A student who thinks the amount of solid affects the position of a dissolution equilibrium picks this. Solid Ca(OH)₂ does not appear in Ksp = [Ca²⁺][OH⁻]², so its mass is not needed, provided some solid was present when the solution was saturated.
BThe total volume of saturated solution made A student who thinks ΔG° depends on the size of the sample picks this. [OH⁻] is the same in any portion of the saturated solution, and ΔG° is per mole of reaction, so only the measured volume that is titrated is needed, and the plan already includes it.
CThe time taken for the solid to stop dissolving A student who thinks the speed of a process is needed to find its equilibrium constant picks this. How quickly the solid dissolved has no effect on Ksp or ΔG°, which describe the saturated solution once equilibrium is reached.
DThe temperature of the saturated solutionCorrect ΔG° = −RT ln K. The titration gives [OH⁻], and so [Ca²⁺] = ½[OH⁻] and Ksp, but the calculation also needs T, the temperature in kelvins at which the solution was saturated and Ksp applies.
For a gas-phase reaction in which one reactant molecule forms one product molecule, ΔG° = −1.0 kJ/molrxn at 298 K. At 298 K, RT = 2.5 kJ/mol. A student claims that at equilibrium at 298 K nearly all of the reactant will have been converted to product. Which evaluation of the claim is correct?
Answer and reasoning
AThe claim is not supported: ΔG° is close to zero, so K is close to 1Correct −ΔG°/RT = 1.0/2.5 = 0.40, so K = e0.40 ≈ 1.5. ΔG° is close to zero (its magnitude is smaller than RT), so K is close to 1, and the equilibrium mixture contains comparable amounts of reactant and product (about 60% product), not nearly all product.
BThe claim is supported: ΔG° is negative, so the reaction goes to completion A student who thinks any reaction with ΔG° < 0 goes essentially to completion picks this. The sign shows only that K > 1; the size of ΔG° compared with RT shows that K ≈ 1.5, so much of the reactant remains.
CThe claim is not supported: ΔG° is negative, so K is less than 1 A student who matches a negative ΔG° with a K less than 1 picks this. ΔG° < 0 gives K > 1 (here about 1.5); the claim fails because K is only slightly greater than 1, not because the reactant is favored.
DThe claim is not supported: ΔG° is not zero, so equilibrium is not reached A student who thinks ΔG° must be zero at equilibrium picks this. The system does reach equilibrium, with K = e0.40 ≈ 1.5; ΔG° determines the value of K, not whether equilibrium can be reached.
At 450 K, ΔG° = −20.0 kJ/molrxn for reaction 1. At the same temperature, the equilibrium constant of reaction 2 is exactly 10 times the equilibrium constant of reaction 1. What is ΔG° for reaction 2 at 450 K?
Answer and reasoning
A−25.7 kJ/molrxn A student who uses T = 298 K because of the ° symbol picks this: RT ln 10 = (8.314)(298)(2.303) J/mol = 5.7 kJ/mol. The temperature in the equation is the temperature of the system, 450 K, which gives a change of 8.6 kJ/mol.
B−23.7 kJ/molrxn A student who treats ln as log picks this: log 10 = 1, so the change is taken as RT = (8.314)(450) J/mol = 3.7 kJ/mol. The equation uses the natural logarithm, and ln 10 = 2.303.
C−11.4 kJ/molrxn A student who thinks ΔG° and K rise and fall together picks this, adding RT ln 10 = 8.6 kJ/mol for the larger K. The negative sign in ΔG° = −RT ln K means a larger K gives a more negative ΔG°.
D−28.6 kJ/molrxnCorrect Because ΔG° = −RT ln K, multiplying K by 10 changes ΔG° by −RT ln 10 = −(8.314)(450)(2.303) J/mol = −8.6 kJ/mol, so ΔG° for reaction 2 is −20.0 − 8.6 = −28.6 kJ/molrxn.
Working ΔG°₂ = −RT ln K₂ = −RT ln(10 K₁) = −RT ln K₁ − RT ln 10 = ΔG°₁ − RT ln 10. RT ln 10 = (8.314 J/(mol·K))(450 K)(2.303) = 8.61 × 10³ J/mol = 8.6 kJ/mol. ΔG°₂ = −20.0 − 8.6 = −28.6 kJ/molrxn. A larger K goes with a more negative ΔG°.
At 298 K, the reaction X(g) ⇌ Y(g) has K = 5.0 and ΔG° = −4.0 kJ/molrxn. The equation is then rewritten as 2 X(g) ⇌ 2 Y(g). How do K and ΔG° for the rewritten equation at 298 K compare with the original values?
Answer and reasoning
AK and ΔG° are both doubled A student who takes K to be proportional to −ΔG° picks this, doubling K to 10 when ΔG° doubles. ΔG° is proportional to ln K, not to K, so doubling ΔG° squares K: −RT ln 10 is −5.7 kJ/molrxn, not −8.0 kJ/molrxn.
BK and ΔG° are both unchanged A student who thinks K at a given temperature is the same however the equation is written picks this, and keeps ΔG° = −RT ln K the same too. K belongs to the equation as written: doubling the coefficients squares K and doubles ΔG°.
CK is squared and ΔG° is unchanged A student who treats ΔG° as a fixed property of the reaction, while squaring K for the doubled equation, picks this. With K = 25, ΔG° = −RT ln 25 = −8.0 kJ/molrxn, twice the original value.
DK is squared and ΔG° is doubledCorrect ΔG° is per mole of reaction as written, so it doubles to −8.0 kJ/molrxn. Since ln K = −ΔG°/RT, doubling ΔG° doubles ln K, which squares K: K = (5.0)² = 25, and −RT ln 25 = −8.0 kJ/molrxn.
Working ΔG° is per mole of reaction as written, so doubling the coefficients doubles it: ΔG° = −8.0 kJ/molrxn. Then K = e−ΔG°/RT = e2 ln 5.0 = (5.0)² = 25. Check: −RT ln 25 = −(8.314)(298)(3.22) J/mol = −8.0 kJ/molrxn. Doubling ΔG° doubles ln K, which squares K.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
