AP Chemistry Practice Test 8

60 original exam-style multiple-choice questions with answers & explanations — covering all nine units of the AP Chemistry course.

Note: These are original practice questions written in the AP Chemistry exam style — not reproduced College Board questions. Use them to test your readiness for the real AP Chemistry exam.

📱 How to use: Try each question first, then click “Show Answer” to check. Covers all 9 AP units: atomic structure, compound structure, intermolecular forces, reactions, kinetics, thermodynamics, equilibrium, acids & bases, and applications of thermodynamics.

⚛ Unit 1: Atomic Structure & Properties

Q1Naturally occurring chlorine consists of two isotopes: ^35Cl (mass 34.97 u, 75.77% abundance) and ^37Cl (mass 36.97 u, 24.23% abundance). The average atomic mass of chlorine is closest to

(A) 34.97 u
(B) 36.97 u
(C) 35.45 u
(D) 35.97 u
Show Answer

Correct: (C)

0.7577(34.97) + 0.2423(36.97) = 26.50 + 8.96 = 35.45 u.

Q2The ground-state electron configuration of a copper atom is

(A) [Ar]4s^1 3d^10
(B) [Ar]4s^2 3d^9
(C) [Ar]3d^11
(D) [Ar]4s^2 3d^10
Show Answer

Correct: (A)

Copper is an exception to the expected filling order; a completely filled 3d subshell gives [Ar]4s^1 3d^10.

Q3The photoelectron spectrum of an element shows four peaks with relative heights of 2, 2, 6, and 1, in order of decreasing binding energy. The element is

(A) magnesium
(B) aluminum
(C) potassium
(D) sodium
Show Answer

Correct: (D)

The peaks correspond to 1s^2, 2s^2, 2p^6, and 3s^1, a total of 11 electrons, which is sodium.

Q4Which of the following elements has the highest first ionization energy?

(A) Al
(B) Si
(C) Na
(D) Mg
Show Answer

Correct: (B)

First ionization energy generally increases across period 3, and Si lies farthest to the right among these elements (the dip at Al makes Al lower than Mg).

Q5A mass spectrum of magnesium shows three peaks: m/z 24 (79.0% abundance), m/z 25 (10.0% abundance), and m/z 26 (11.0% abundance). Based only on these data, the average atomic mass of magnesium is closest to

(A) 24.0 u
(B) 25.0 u
(C) 24.3 u
(D) 24.7 u
Show Answer

Correct: (C)

0.79(24) + 0.10(25) + 0.11(26) = 18.96 + 2.50 + 2.86 = 24.3 u.

Q6The ground-state electron configuration of the Ni^2+ ion is

(A) [Ar]3d^8
(B) [Ar]4s^2 3d^6
(C) [Ar]4s^1 3d^7
(D) [Ar]3d^10
Show Answer

Correct: (A)

Ni is [Ar]4s^2 3d^8; Ni^2+ forms by removing the two 4s electrons, leaving [Ar]3d^8.

Q7Which of the following atoms has the smallest atomic radius?

(A) S
(B) O
(C) Se
(D) Te
Show Answer

Correct: (B)

Atomic radius increases down a group, so oxygen, at the top of group 16, is the smallest.

🔗 Unit 2: Molecular & Ionic Compound Structure

Q8In the Lewis structure of NH3, the nitrogen atom has

(A) three bonding pairs and no lone pairs
(B) two bonding pairs and two lone pairs
(C) three bonding pairs and one lone pair
(D) four bonding pairs and one lone pair
Show Answer

Correct: (C)

Nitrogen forms three N-H bonds and retains one lone pair, giving a trigonal pyramidal molecular geometry.

Q9How many equivalent resonance structures can be drawn for SO2 in which formal charges are minimized (one S=O double bond and one S-O single bond in each structure)?

(A) 1
(B) 3
(C) 4
(D) 2
Show Answer

Correct: (D)

The double bond can be placed on either of the two equivalent oxygen atoms, giving 2 equivalent resonance structures.

Q10In carbon monoxide, CO, drawn with a triple bond and one lone pair on each atom, the formal charge on the carbon atom is

(A) 0
(B) -1
(C) +1
(D) -2
Show Answer

Correct: (B)

Formal charge = 4 – (2 nonbonding electrons + 6 bonding electrons/2) = 4 – 5 = -1.

Q11The molecular geometry of BrF5 is

(A) square pyramidal
(B) trigonal bipyramidal
(C) octahedral
(D) pentagonal planar
Show Answer

Correct: (A)

BrF5 has 6 electron domains (5 bonds, 1 lone pair) in an octahedral arrangement; the lone pair makes the molecular shape square pyramidal.

Q12The hybridization of the sulfur atom in SF6 is

(A) sp^3
(B) sp^3d
(C) sp^3d^2
(D) sp^2
Show Answer

Correct: (C)

Sulfur forms six bonding domains in an octahedral arrangement, which requires sp^3d^2 hybridization.

Q13Which of the following compounds has the greatest lattice energy?

(A) LiF
(B) NaCl
(C) KBr
(D) MgO
Show Answer

Correct: (D)

Lattice energy is greatest for small, highly charged ions; Mg^2+ and O^2- carry charges of +2 and -2, so MgO far exceeds the +1/-1 salts.

Q14Which of the following bonds is the shortest?

(A) a C-C single bond in ethane
(B) a C≡C triple bond in ethyne
(C) a C=C double bond in ethene
(D) an N-N single bond in hydrazine
Show Answer

Correct: (B)

Bond length decreases as bond order increases, so a triple bond is shorter than double and single bonds between similar atoms.

💧 Unit 3: Intermolecular Forces & Properties

Q15Which of the following substances has the highest normal boiling point?

(A) H2O
(B) H2S
(C) H2Se
(D) H2Te
Show Answer

Correct: (A)

Water hydrogen bonds strongly, giving it an anomalously high boiling point even though it has the lowest molar mass of the group 16 hydrides.

Q16Which of the following molecules experiences only London dispersion forces as its intermolecular force?

(A) NH3
(B) CH3Cl
(C) HF
(D) CH4
Show Answer

Correct: (D)

CH4 is nonpolar and has no hydrogen bonded to N, O, or F, so only dispersion forces act between its molecules.

Q17How much heat is required to vaporize 2.00 mol of water at 100°C? (ΔHvap of water = 40.7 kJ/mol)

(A) 40.7 kJ
(B) 81.4 kJ
(C) 20.4 kJ
(D) 122 kJ
Show Answer

Correct: (B)

q = 2.00 mol × 40.7 kJ/mol = 81.4 kJ.

Q18Which of the following liquids is expected to have the greatest surface tension?

(A) diethyl ether
(B) acetone
(C) water
(D) hexane
Show Answer

Correct: (C)

Surface tension increases with the strength of intermolecular forces; water's extensive hydrogen bonding gives it the greatest surface tension of these liquids.

Q19Which of the following solids is a good conductor of electricity in the solid state?

(A) diamond
(B) sulfur
(C) sodium chloride
(D) graphite
Show Answer

Correct: (D)

Graphite has delocalized pi electrons that move freely within its layers; diamond, sulfur, and solid NaCl have no mobile charge carriers.

Q20What is the molarity of a solution made by dissolving 5.85 g of NaCl (molar mass 58.5 g/mol) in enough water to make 500 mL of solution?

(A) 0.200 M
(B) 0.100 M
(C) 0.400 M
(D) 11.7 M
Show Answer

Correct: (A)

mol NaCl = 5.85 / 58.5 = 0.100 mol; molarity = 0.100 mol / 0.500 L = 0.200 M.

⚖ Unit 4: Chemical Reactions

Q21When the equation Fe2O3 + CO → Fe + CO2 is balanced with the smallest whole-number coefficients, the coefficient of CO is

(A) 1
(B) 2
(C) 3
(D) 6
Show Answer

Correct: (C)

Fe2O3 + 3 CO → 2 Fe + 3 CO2.

Q22The oxidation number of manganese in KMnO4 is

(A) +7
(B) +4
(C) +2
(D) +6
Show Answer

Correct: (A)

With K at +1 and four oxygens at -2 (total -8), manganese must be +7 for the neutral compound to sum to zero.

Q23Solutions of Ba(OH)2 and H2SO4 are mixed, producing a precipitate and water. The correct net ionic equation for the reaction is

(A) H^+(aq) + OH^-(aq) → H2O(l)
(B) Ba^2+(aq) + SO4^2-(aq) → BaSO4(s)
(C) Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + 2 H2O(l)
(D) Ba^2+(aq) + 2 OH^-(aq) + 2 H^+(aq) + SO4^2-(aq) → BaSO4(s) + 2 H2O(l)
Show Answer

Correct: (D)

Both the precipitate BaSO4 and water form, so the net ionic equation must include Ba^2+, SO4^2-, H^+, and OH^-; choices A and B each show only half of the reaction, and C is the molecular equation.

Q24Calcium carbonate decomposes on heating: CaCO3(s) → CaO(s) + CO2(g). The mass of CO2 produced when 10.0 g of CaCO3 (molar mass 100 g/mol) decomposes completely is

(A) 10.0 g
(B) 4.40 g
(C) 2.20 g
(D) 44.0 g
Show Answer

Correct: (B)

10.0 g / 100 g/mol = 0.100 mol CaCO3 produces 0.100 mol CO2 = 0.100 × 44.0 = 4.40 g.

Q25For the reaction 2 Al + 3 Cl2 → 2 AlCl3, a mixture of 1.0 mol Al and 1.0 mol Cl2 reacts completely. The limiting reactant is

(A) Al
(B) neither; both are consumed exactly
(C) Cl2
(D) AlCl3
Show Answer

Correct: (C)

1.0 mol Al would require 1.5 mol Cl2, but only 1.0 mol Cl2 is present, so Cl2 is limiting.

Q26A synthesis has a theoretical yield of 30.0 g, but only 27.0 g of product is actually isolated. The percent yield is

(A) 27%
(B) 111%
(C) 97%
(D) 90%
Show Answer

Correct: (D)

Percent yield = (27.0 / 30.0) × 100% = 90%.

Q27A sample of H2SO4 is exactly neutralized by 20.0 mL of 0.150 M NaOH. The number of moles of H2SO4 in the sample is

(A) 0.00150 mol
(B) 0.00300 mol
(C) 0.00600 mol
(D) 0.00750 mol
Show Answer

Correct: (A)

mol NaOH = 0.150 × 0.0200 = 0.00300 mol; because H2SO4 + 2 NaOH, mol H2SO4 = 0.00300 / 2 = 0.00150 mol.

⏱ Unit 5: Kinetics

Q28Initial-rate experiments for the reaction A + B → products show that doubling [A] leaves the rate unchanged, while doubling [B] doubles the rate. The rate law is

(A) Rate = k[A][B]
(B) Rate = k[B]
(C) Rate = k[A]
(D) Rate = k[A]^2[B]
Show Answer

Correct: (B)

The rate is independent of [A] (zero order) and doubles when [B] doubles (first order in B).

Q29A radioactive isotope has a half-life of 5.0 days. What fraction of an original sample remains after 15 days?

(A) 1/2
(B) 1/4
(C) 1/16
(D) 1/8
Show Answer

Correct: (D)

15 days is three half-lives, so the fraction remaining is (1/2)^3 = 1/8.

Q30The units of the rate constant for a first-order reaction are

(A) s^-1
(B) M s^-1
(C) M^-1 s^-1
(D) M^-2 s^-1
Show Answer

Correct: (A)

For rate = k[A], k = rate/[A] has units of (M/s)/M = s^-1.

Q31A proposed mechanism for a reaction is: Step 1: A2 ⇌ 2 A (fast equilibrium); Step 2: A + B → products (slow). The rate law predicted by this mechanism is

(A) Rate = k[A2][B]
(B) Rate = k[A]^2[B]
(C) Rate = k[A2]^1/2[B]
(D) Rate = k[B]
Show Answer

Correct: (C)

The slow step gives Rate = k[A][B]; from the fast equilibrium, [A]^2 is proportional to [A2], so [A] is proportional to [A2]^1/2, giving Rate = k'[A2]^1/2[B].

Q32According to collision theory, increasing the concentration of reactants increases the reaction rate because

(A) the activation energy decreases
(B) collisions between reactant particles occur more frequently
(C) the temperature of the system rises
(D) more product is formed at equilibrium
Show Answer

Correct: (B)

A higher concentration means more particles per unit volume, so collisions occur more often and the rate increases.

Q33When the temperature of a reaction mixture is raised from 300 K to 310 K, the rate roughly doubles, mainly because

(A) the activation energy is cut in half
(B) the equilibrium constant doubles
(C) the concentrations of the reactants increase
(D) a much larger fraction of the molecules have kinetic energy greater than the activation energy
Show Answer

Correct: (D)

A modest temperature rise greatly increases the fraction of molecules able to overcome the activation energy barrier; the activation energy itself does not change.

Q34Adding a catalyst to a reaction mixture does NOT change the

(A) ΔH of the reaction
(B) activation energy of the forward reaction
(C) rate of the forward reaction
(D) mechanism of the reaction
Show Answer

Correct: (A)

A catalyst changes the pathway and lowers the activation energy, but the energies of the reactants and products, and therefore ΔH, are unchanged.




🔥 Unit 6: Thermodynamics

Q35A reaction in a calorimeter causes the temperature of 200 g of water to increase by 10.0°C. The heat released by the reaction is (specific heat of water = 4.18 J/g·°C)

(A) 836 J
(B) 4,180 J
(C) 8,360 J
(D) 83,600 J
Show Answer

Correct: (C)

q = mcΔT = (200)(4.18)(10.0) = 8,360 J absorbed by the water, so the reaction released 8,360 J.

Q36Given: N2(g) + O2(g) → 2 NO(g), ΔH = +180.5 kJ; and 2 NO(g) + O2(g) → 2 NO2(g), ΔH = -114.1 kJ. The ΔH for the reaction NO(g) + 1/2 O2(g) → NO2(g) is

(A) +66.4 kJ
(B) -57.1 kJ
(C) -114.1 kJ
(D) +33.2 kJ
Show Answer

Correct: (B)

The target reaction is exactly one-half of the second given reaction, so ΔH = -114.1 / 2 = -57.1 kJ.

Q37Using average bond enthalpies (N≡N = 941 kJ/mol, H-H = 436 kJ/mol, N-H = 391 kJ/mol), the estimated ΔH for N2(g) + 3 H2(g) → 2 NH3(g) is

(A) +97 kJ
(B) -2,346 kJ
(C) -4,595 kJ
(D) -97 kJ
Show Answer

Correct: (D)

Bonds broken: 941 + 3(436) = 2,249 kJ; bonds formed: 6(391) = 2,346 kJ; ΔH = 2,249 – 2,346 = -97 kJ.

Q38Given standard enthalpies of formation ΔHf°(SO2, g) = -296.8 kJ/mol and ΔHf°(SO3, g) = -395.7 kJ/mol, the ΔH° for the reaction 2 SO2(g) + O2(g) → 2 SO3(g) is

(A) -197.8 kJ
(B) +197.8 kJ
(C) -98.9 kJ
(D) -692.5 kJ
Show Answer

Correct: (A)

ΔH = 2(-395.7) – [2(-296.8) + 0] = -791.4 + 593.6 = -197.8 kJ.

Q39Which of the following processes has ΔS > 0 for the system?

(A) condensation of water vapor
(B) freezing of benzene
(C) sublimation of dry ice, CO2(s) → CO2(g)
(D) precipitation of AgCl when AgNO3 and NaCl solutions are mixed
Show Answer

Correct: (C)

Sublimation converts an ordered solid directly into a gas, greatly increasing the disorder of the system; the other processes decrease entropy.

Q40How many grams of water can be heated by 2.0°C using 418 J of heat? (specific heat of water = 4.18 J/g·°C)

(A) 25 g
(B) 50 g
(C) 100 g
(D) 200 g
Show Answer

Correct: (B)

m = q / (cΔT) = 418 / (4.18 × 2.0) = 50 g.

Q41Steam at 100°C causes more severe burns than liquid water at 100°C mainly because

(A) steam has a higher temperature than boiling water
(B) steam is less dense than liquid water
(C) the molecules in steam move faster than those in liquid water
(D) steam releases a large amount of heat when it condenses on the skin
Show Answer

Correct: (D)

Condensation of steam releases the heat of vaporization (about 2,260 J/g) in addition to the heat released as the liquid cools, delivering far more energy to the skin.

⇆ Unit 7: Equilibrium

Q42For the equilibrium CaCO3(s) ⇌ CaO(s) + CO2(g), the equilibrium constant expression Kc is

(A) Kc = [CO2]
(B) Kc = [CaO][CO2] / [CaCO3]
(C) Kc = [CaCO3] / ([CaO][CO2])
(D) Kc = [CO2]^2
Show Answer

Correct: (A)

Pure solids are omitted from equilibrium constant expressions, leaving only the gaseous CO2.

Q43For the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g), Kc = 0.50 at a certain temperature. A flask initially contains 1.0 M PCl5 and no products. The equilibrium concentration of PCl5 is

(A) 1.0 M
(B) 0.25 M
(C) 0.75 M
(D) 0.50 M
Show Answer

Correct: (D)

Setting x^2 / (1.0 – x) = 0.50 gives x^2 + 0.50x – 0.50 = 0, so x = 0.50 M and [PCl5] = 1.0 – 0.50 = 0.50 M.

Q44For the equilibrium 2 NO(g) + O2(g) ⇌ 2 NO2(g), increasing the pressure by decreasing the volume of the container will

(A) shift the equilibrium toward reactants
(B) shift the equilibrium toward products
(C) have no effect on the equilibrium position
(D) decrease the value of Kc
Show Answer

Correct: (B)

The reactant side has 3 mol of gas and the product side has 2 mol; increased pressure shifts the equilibrium toward the side with fewer moles of gas, the products. Kc itself does not change.

Q45The Ksp of Ca(OH)2 is 5.5 × 10^-6. The molar solubility of Ca(OH)2 in pure water is closest to

(A) 5.5 × 10^-6 M
(B) 1.8 × 10^-3 M
(C) 1.1 × 10^-2 M
(D) 2.3 × 10^-2 M
Show Answer

Correct: (C)

For Ca(OH)2, Ksp = [Ca^2+][OH^-]^2 = (s)(2s)^2 = 4s^3, so s = (5.5 × 10^-6 / 4)^(1/3) = 1.1 × 10^-2 M.

Q46A reaction mixture is analyzed and the reaction quotient Q is found to be less than the equilibrium constant K. Which statement is true?

(A) The reaction will proceed toward products until Q = K
(B) The reaction will proceed toward reactants until Q = K
(C) The system is already at equilibrium
(D) The forward reaction rate is zero
Show Answer

Correct: (A)

When Q < K, reactants are in excess relative to equilibrium, so the net reaction proceeds forward, forming products until Q = K.

Q47The molar solubility of AgCl in a solution that already contains 0.10 M NaCl is lower than its solubility in pure water. The best explanation is

(A) the temperature of the solution is lower
(B) the Ksp of AgCl increases in the presence of NaCl
(C) the ionic strength changes the density of the solution
(D) the common ion effect: the Cl^- already present shifts the dissolution equilibrium toward solid AgCl
Show Answer

Correct: (D)

Added Cl^-, a common ion, shifts AgCl(s) ⇌ Ag^+(aq) + Cl^-(aq) to the left by Le Chatelier's principle, reducing the solubility; Ksp is unchanged.

Q48For a reaction in which Δn(gas) = +1, Kc = 0.10 at 300 K. The value of Kp is closest to (R = 0.0821 L·atm/mol·K)

(A) 0.10
(B) 2.5
(C) 0.0041
(D) 246
Show Answer

Correct: (B)

Kp = Kc(RT)^Δn = 0.10 × (0.0821 × 300)^1 = 0.10 × 24.6 = 2.5.

🧫 Unit 8: Acids & Bases

Q49The pH of a 0.0010 M NaOH solution is

(A) 3.00
(B) 10.00
(C) 11.00
(D) 1.00
Show Answer

Correct: (C)

[OH^-] = 0.0010 M, so pOH = 3.00 and pH = 14.00 – 3.00 = 11.00.

Q50The pH of a 0.10 M NH3 solution (Kb = 1.8 × 10^-5) is closest to

(A) 11.13
(B) 2.87
(C) 13.00
(D) 8.00
Show Answer

Correct: (A)

[OH^-] = sqrt(Kb × C) = sqrt(0.10 × 1.8 × 10^-5) = 1.3 × 10^-3 M; pOH = 2.87, so pH = 14.00 – 2.87 = 11.13.

Q51A buffer solution is prepared with 0.10 M HF and 0.10 M NaF. Given that the pKa of HF is 3.17, the pH of the buffer is

(A) 2.87
(B) 3.47
(C) 7.00
(D) 3.17
Show Answer

Correct: (D)

pH = pKa + log([F^-]/[HF]) = 3.17 + log(1.0) = 3.17.

Q52A weak acid is titrated with a strong base. At the point where exactly one-half of the acid has been neutralized, which statement is true?

(A) The pH equals 7.00
(B) The pH equals the pKa of the acid
(C) The pH equals 14.00 – pKa of the acid
(D) The solution contains only the conjugate base
Show Answer

Correct: (B)

At half-neutralization [HA] = [A^-], so the Henderson-Hasselbalch equation gives pH = pKa + log(1) = pKa.

Q53The conjugate acid of HCO3^- is

(A) CO3^2-
(B) CO2
(C) H2CO3
(D) OH^-
Show Answer

Correct: (C)

Adding one proton to HCO3^- gives H2CO3, its conjugate acid.

Q54The pH of a 0.010 M H2SO4 solution is closest to (Ka2 for HSO4^- = 1.2 × 10^-2)

(A) 2.00
(B) 1.70
(C) 1.00
(D) 1.84
Show Answer

Correct: (D)

The first ionization is complete, giving 0.010 M H^+; the second ionization (Ka2 = 0.012) contributes about 0.0045 M more, for [H^+] = 0.0145 M and pH = 1.84. (Assuming both protons ionize fully would give pH = 1.70, and first ionization only would give 2.00.)

🔋 Unit 9: Applications of Thermodynamics

Q55A reaction has ΔG° = -50 kJ/mol and ΔH° = -120 kJ/mol at 300 K. The value of ΔS° for the reaction is

(A) -233 J/mol·K
(B) +233 J/mol·K
(C) -567 J/mol·K
(D) -0.23 J/mol·K
Show Answer

Correct: (A)

ΔS = (ΔH – ΔG) / T = (-120 – (-50)) / 300 = -0.233 kJ/mol·K = -233 J/mol·K.

Q56A proposed redox reaction has E°cell = -0.50 V under standard conditions. Which statement about the reaction as written is correct?

(A) It is spontaneous as written
(B) Its equilibrium constant K is greater than 1
(C) ΔG° is positive and the reaction is nonspontaneous as written
(D) It operates as a galvanic (voltaic) cell
Show Answer

Correct: (C)

A negative E°cell means ΔG° = -nFE° is positive, so the reaction is nonspontaneous as written (it would run as an electrolytic process with added energy) and K < 1.

Q57What is E°cell for the reaction 2 Ag^+(aq) + Cd(s) → 2 Ag(s) + Cd^2+(aq)? (E° for Ag^+/Ag = +0.80 V; E° for Cd^2+/Cd = -0.40 V)

(A) +0.40 V
(B) +1.20 V
(C) -1.20 V
(D) +2.00 V
Show Answer

Correct: (B)

Ag^+ is reduced and Cd is oxidized: E°cell = 0.80 – (-0.40) = +1.20 V. (Potentials are not multiplied by the stoichiometric coefficients.)

Q58Silver is plated from an AgNO3 solution using a current of 5.00 A for 9,650 s. The mass of Ag deposited is closest to (F = 96,500 C/mol; molar mass of Ag = 107.9 g/mol)

(A) 27.0 g
(B) 5.39 g
(C) 10.8 g
(D) 53.9 g
Show Answer

Correct: (D)

Charge = 5.00 × 9,650 = 48,250 C = 0.500 mol e^-. Since Ag^+ + e^- → Ag, 0.500 mol Ag forms, with mass 0.500 × 107.9 = 53.9 g.

Q59A reaction is spontaneous at all temperatures when

(A) ΔH < 0 and ΔS > 0
(B) ΔH > 0 and ΔS < 0
(C) ΔH < 0 and ΔS < 0
(D) ΔH > 0 and ΔS > 0
Show Answer

Correct: (A)

With ΔH negative and ΔS positive, ΔG = ΔH – TΔS is negative at every temperature.

Q60If ΔG° for a reaction is negative at 298 K, then its equilibrium constant K is

(A) less than 1
(B) equal to 1
(C) greater than 1
(D) equal to 0
Show Answer

Correct: (C)

ΔG° = -RT ln K; a negative ΔG° means ln K > 0, so K > 1 and products are favored at equilibrium.