ACS Gen Chem Practice Questions (2026) — Free Sample Questions with Answers

Answer: C

1. 1. λ = 532 nm = 5.32 × 10⁻⁷ m
2. 2. E(photon) = hc/λ = (6.626 × 10⁻³⁴)(2.998 × 10⁸)/(5.32 × 10⁻⁷) = 3.73 × 10⁻¹⁹ J
3. 3. E(mol) = 3.73 × 10⁻¹⁹ J × 6.022 × 10²³ mol⁻¹ = 2.25 × 10⁵ J/mol
4. 4. 2.25 × 10⁵ J/mol ÷ 1000 = 225 kJ/mol

3.73 × 10⁻²² kJ/mol: This is the energy of a single photon (3.73 × 10⁻¹⁹ J, here in kJ). It was never multiplied by Avogadro's number.

7.50 × 10⁻⁷ kJ/mol: Uses E = h/λ and leaves out c. The photon energy is E = hc/λ.

225 kJ/mol: Correct. E = hc/λ = (6.626 × 10⁻³⁴ J·s)(2.998 × 10⁸ m/s)/(5.32 × 10⁻⁷ m) = 3.73 × 10⁻¹⁹ J per photon. × 6.022 × 10²³ mol⁻¹ = 2.25 × 10⁵ J/mol = 225 kJ/mol.

2.25 × 10⁵ kJ/mol: The number is in J/mol but is labelled kJ/mol. It still needs dividing by 1000.

E = hc/λ gives the energy of one photon. Multiply by N_A for a mole of photons, and convert nm to m first.

Answer: D

$\mathrm{XO}$: Fits an element with 2 valence electrons, which would show its big jump between IE₂ and IE₃.

${\mathrm{X}}_2\mathrm{O}$: Fits a group 1 element, whose big jump would come right after IE₁.

$\mathrm{X}{\mathrm{O}}_2$: Fits 4 valence electrons (X⁴⁺), which would put the jump between IE₄ and IE₅. Here the jump comes after IE₃.

${\mathrm{X}}_2{\mathrm{O}}_3$: Correct. The large jump comes between IE₃ and IE₄. That means X has 3 valence electrons (it is Al), forms X³⁺, and combines with O²⁻ as X₂O₃.

A very large jump in successive ionization energies marks the first core electron. The number of electrons removed before the jump is the number of valence electrons.

Answer: D

Structure I, because every atom has two bonds and the bonding is symmetric: In I the −1 formal charge sits on N, which is less electronegative than O. Symmetry of bond orders is not the criterion.

Structure II, because oxygen has a triple bond and nitrogen has three lone pairs: II has the largest formal charges (+1 on O, −2 on N) and puts a positive charge on the most electronegative atom. It contributes least.

All three contribute equally because they have the same number of electrons: Resonance structures always have the same electron count. Their contributions differ according to their formal-charge distributions.

Structure III, because the −1 formal charge is on oxygen, the most electronegative atom: Correct. Formal charges (O, C, N): I = 0, 0, −1; II = +1, 0, −2; III = −1, 0, 0. Structures I and III both keep charges small, but III puts the negative charge on the more electronegative O.

Formal charge = valence e⁻ − nonbonding e⁻ − ½(bonding e⁻). Prefer the structure with the smallest formal charges and any negative charge on the most electronegative atom.

Answer: A

sp², sp², sp: Correct. C1 and C2 each have three electron domains (the C=C counts as one), so they are sp². C3 has two domains (C–C and C≡N), so it is sp.

sp², sp², sp²: Misses that the triple bond leaves C3 with only two electron domains, which makes it linear and sp.

sp, sp, sp²: Counts each bond line (double = 2) as a domain and reverses the result. A multiple bond counts as one domain.

sp³, sp², sp: Treats C1 as having four domains. Its two H atoms and the double bond make three.

Hybridization follows from the number of electron domains: 2 = sp, 3 = sp², 4 = sp³. A double or triple bond counts as one domain.

Answer: C

1. 1. The higher plateau (constant T = 60 °C) is boiling.
2. 2. Heat absorbed on that plateau = 22.5 kJ − 7.5 kJ = 15.0 kJ
3. 3. ΔH_vap = 15.0 kJ ÷ 0.500 mol = 30.0 kJ/mol

6.00 kJ/mol: Uses the lower plateau (−20 °C, 3.0 kJ), which is melting. Vaporization is the higher-temperature plateau.

15.0 kJ/mol: This is the heat absorbed by the whole sample (22.5 − 7.5 kJ), not divided by 0.500 mol.

30.0 kJ/mol: Correct. The plateau at 60 °C (the boiling point) runs from 7.5 to 22.5 kJ, so q = 15.0 kJ, and ΔH_vap = 15.0 kJ ÷ 0.500 mol = 30.0 kJ/mol.

45.0 kJ/mol: Divides the heat value at the end of the plateau (22.5 kJ) by 0.500 mol instead of using the length of the plateau.

On a heating curve, the plateaus are phase changes. The length of a plateau (in heat units) divided by moles gives that phase change's molar enthalpy.

Answer: B

1. 1. The mass stops increasing at 0.369 g (trials 4–6), so in those trials Pb²⁺ is the limiting reactant and is fully precipitated.
2. 2. n(PbI₂) = 0.369 g ÷ 461.0 g/mol = 8.00 × 10⁻⁴ mol = n(Pb²⁺) in 20.0 mL
3. 3. M(Pb(NO₃)₂) = 8.00 × 10⁻⁴ mol ÷ 0.0200 L = 0.0400 M
4. 4. Check: at 8.0 mL, n(I⁻) = 0.0080 L × 0.200 M = 1.60 × 10⁻³ mol, exactly 2 × n(Pb²⁺).

0.0286 M: Divides by the combined volume (20.0 + 8.0 mL). The lead ions came from the original 20.0 mL sample only.

0.0400 M: Correct. The mass levels off at 0.369 g once all $\mathrm{P}{\mathrm{b}}^{2+}$ has precipitated: 0.369 g ÷ 461.0 g/mol = 8.00 × 10⁻⁴ mol $\mathrm{Pb}{\mathrm{I}}_2$ = mol $\mathrm{P}{\mathrm{b}}^{2+}$; ÷ 0.0200 L = 0.0400 M. (Check: 8.0 mL × 0.200 M = 1.60 × 10⁻³ mol ${\mathrm{I}}^{-}$ = 2 × 8.00 × 10⁻⁴.)

0.0600 M: Uses the largest KI volume (12.0 mL). Beyond 8.0 mL the mass no longer increases, so the extra KI is in excess and does not react.

0.0800 M: Uses a 1:1 ratio of ${\mathrm{I}}^{-}$ to $\mathrm{P}{\mathrm{b}}^{2+}$ (0.00160 mol ÷ 0.0200 L); each $\mathrm{P}{\mathrm{b}}^{2+}$ needs 2 ${\mathrm{I}}^{-}$.

When precipitate mass levels off, the fixed reagent has been consumed; its amount equals the moles of precipitate (adjusted by the formula ratio).

Answer: A

4 $\mathrm{N}{\mathrm{H}}_3$; ${\mathrm{N}}_2$ left over: Correct. The drawing has 4 ${\mathrm{N}}_2$ and 6 ${\mathrm{H}}_2$. 4 ${\mathrm{N}}_2$ would need 12 ${\mathrm{H}}_2$, so ${\mathrm{H}}_2$ is limiting: 6 ${\mathrm{H}}_2$ react with 2 ${\mathrm{N}}_2$ to give 4 $\mathrm{N}{\mathrm{H}}_3$, and 2 ${\mathrm{N}}_2$ remain.

2 $\mathrm{N}{\mathrm{H}}_3$; ${\mathrm{N}}_2$ left over: Uses 3 ${\mathrm{H}}_2$ → 1 $\mathrm{N}{\mathrm{H}}_3$, dropping the coefficient 2 on $\mathrm{N}{\mathrm{H}}_3$.

6 $\mathrm{N}{\mathrm{H}}_3$; ${\mathrm{N}}_2$ left over: Uses a 1:1 ratio of ${\mathrm{H}}_2$ to $\mathrm{N}{\mathrm{H}}_3$; 6 $\mathrm{N}{\mathrm{H}}_3$ would contain 18 H atoms, but only 12 are present.

8 $\mathrm{N}{\mathrm{H}}_3$; ${\mathrm{H}}_2$ left over: Assumes the reactant with fewer molecules (${\mathrm{N}}_2$) is limiting. 8 $\mathrm{N}{\mathrm{H}}_3$ would need 12 ${\mathrm{H}}_2$, and only 6 are present.

Limiting reactant: divide the amount of each reactant by its coefficient; the smallest quotient limits. Fewer particles does not mean limiting.

Answer: A

1. 1. Q = [Ni²⁺]/[Ag⁺]² = 0.10/(0.010)² = 1.0 × 10³; log Q = 3.00
2. 2. n = 2
3. 3. E = 1.06 V − (0.0592/2)(3.00) = 1.06 − 0.0888 = 0.97 V

0.97 V: Correct. Q = [Ni²⁺]/[Ag⁺]² = 0.10/(0.010)² = 1.0 × 10³. E = E° − (0.0592/n) log Q = 1.06 − (0.0592/2)(3.00) = 1.06 − 0.089 = 0.97 V.

1.03 V: Leaves out the square on [Ag⁺], so Q = 10 and the correction is only 0.030 V.

1.06 V: This is E°cell. The concentrations are not 1 M, so the Nernst correction applies.

1.15 V: Uses Q inverted (reactants over products), which adds the correction instead of subtracting it.

Nernst equation at 25 °C: E = E° − (0.0592/n) log Q, with coefficients as exponents in Q and solids left out. Q > 1 lowers the cell potential.

Answer: B

The Ag electrode is the anode.: Reverses the roles. The half-reaction with the more positive E° (Ag⁺/Ag) is the reduction, which happens at the cathode.

Electrons flow through the wire from Ni to Ag.: Correct. Ag⁺ has the higher reduction potential, so it is reduced at the Ag cathode while Ni is oxidized at the anode. Electrons leave the anode (Ni) and travel through the wire to the cathode (Ag).

Nitrate ions in the salt bridge move toward the Ag compartment.: Anions migrate toward the anode compartment (Ni), where Ni²⁺ is being produced and positive charge builds up.

The mass of the Ni electrode increases.: Ni(s) is oxidized to Ni²⁺(aq), so the Ni electrode loses mass. It is the Ag electrode that gains mass.

In a galvanic cell, oxidation occurs at the anode and reduction at the cathode (the half-reaction with the larger E°). Electrons flow anode → cathode through the wire; salt-bridge anions move toward the anode.

Answer: D

1. 1. 2 × [S + O₂ → SO₂]: ΔH° = 2(−296.8) = −593.6 kJ
2. 2. Add 2SO₂ + O₂ → 2SO₃: ΔH° = −197.8 kJ. Sum: 2S + 3O₂ → 2SO₃, ΔH° = −791.4 kJ
3. 3. Reverse to match the target: ΔH° = +791.4 kJ

−791.4 kJ: This is ΔH° for the forward reaction 2S + 3O₂ → 2SO₃. The target is the reverse, so the sign must change.

+395.8 kJ: Reverses the first equation's sign but adds the second equation without reversing it (593.6 − 197.8).

+494.6 kJ: Forgets to multiply the first equation by 2; the target needs 2 mol of S.

+791.4 kJ: Correct. Doubling the first equation and adding the second gives 2S + 3O₂ → 2SO₃, ΔH° = 2(−296.8) + (−197.8) = −791.4 kJ. The target is the reverse, so ΔH° = +791.4 kJ.

Hess's law: when an equation is multiplied by n, multiply ΔH by n; when it is reversed, change the sign of ΔH.

Answer: A

1. 1. Linear ln[A] vs t → first order; slope = (−1.623 − (−0.223)) ÷ (80 − 0) s = −0.0175 s⁻¹
2. 2. k = −slope = 0.0175 s⁻¹
3. 3. t½ = 0.693 ÷ 0.0175 s⁻¹ = 39.6 s

39.6 s: Correct. The plot of ln[A] vs t is linear, so the reaction is first order with k = −slope = 1.400 ÷ 80 s = 0.0175 s⁻¹, and t½ = 0.693/k = 39.6 s.

57.1 s: Uses t½ = 1/k. The first-order half-life is ln 2/k = 0.693/k.

71.4 s: Uses the second-order half-life, 1/(k[A]₀). A linear ln[A] plot means first order.

79.2 s: This is two half-lives, the time for [A] to fall to one-quarter of its initial value.

A straight line for ln[A] vs t means first order, with slope = −k and t½ = 0.693/k (independent of concentration).

Answer: A

HY is the weaker acid, and the HY solution has the higher pH.: Correct. Only 1 of 5 HY molecules has ionized (4 HY intact, 1 Y⁻, 1 H₃O⁺), while all 5 HX ionized. Fewer H₃O⁺ in the same volume means a higher pH.

HX and HY are equally strong, because each solution began with 5 acid molecules.: Acid strength depends on the fraction ionized, not on how many molecules were dissolved.

HY is the weaker acid, and the HY solution has the lower pH.: Right acid, wrong pH: the HY solution has fewer H₃O⁺ (1 vs 5), so its pH is higher.

HX is the weaker acid, because no HX molecules remain to donate protons.: Reverses the meaning: every HX molecule has already donated its proton (5 X⁻, 5 H₃O⁺), which is the mark of a strong acid.

A strong acid is (essentially) fully ionized, a weak acid only partly; at equal concentration the weak acid has the lower [H₃O⁺] and the higher pH.

Answer: B

1. 1. Stoichiometry first: NH₃ + H₃O⁺ → NH₄⁺ + H₂O. NH₃: 0.200 − 0.050 = 0.150 mol; NH₄⁺: 0.150 + 0.050 = 0.200 mol.
2. 2. Ka(NH₄⁺) = Kw/Kb = 1.0 × 10⁻¹⁴/1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰; pKa = 9.26.
3. 3. pH = 9.26 + log(0.150/0.200) = 9.26 − 0.12 = 9.13 (9.255 − 0.125 = 9.130).

4.87: Uses pKb (4.74) in place of pKa and the original amounts. That gives the pOH of the original buffer, not the pH.

9.13: Correct. H⁺ converts 0.050 mol NH₃ to NH₄⁺: 0.150 mol NH₃ and 0.200 mol NH₄⁺. pH = 9.26 + log(0.150/0.200) = 9.13.

9.38: The pH before the HCl is added. The strong acid must first react with NH₃.

9.65: Reacts the HCl with NH₄⁺ instead of NH₃ (0.250 mol NH₃, 0.100 mol NH₄⁺). Added acid consumes the base.

For a buffer plus strong acid or base, do the neutralization stoichiometry in moles first, then apply Henderson–Hasselbalch with the new amounts.

Answer: C

1. 1. ICE: [H₂] = [I₂] = 0.500 − x; [HI] = 2x.
2. 2. K = (2x)²/(0.500 − x)² = 49.0, so 2x/(0.500 − x) = 7.00.
3. 3. 2x = 3.50 − 7.00x, so 9.00x = 3.50 and x = 0.389 M.
4. 4. [HI] = 2x = 0.778 M

0.111 M: This is [H₂] (and [I₂]) at equilibrium, 0.500 − 0.389, not [HI].

0.389 M: This is x, the change in [H₂]. [HI] increases by 2x.

0.778 M: Correct. Taking the square root: 2x/(0.500 − x) = 7.00, so x = 0.389 M and [HI] = 2x = 0.778 M.

3.50 M: Neglects x in the denominator (2x/0.500 = 7.00). With K this large, x is not small compared with 0.500 M.

Set up an ICE table with changes in the coefficient ratio. When both sides are perfect squares, take the square root instead of using the quadratic formula.

Answer: C

1. 1. Calibration line: A = (8.00 M⁻¹)c, so c = A ÷ 8.00 M⁻¹
2. 2. Diluted solution: c = 0.280 ÷ 8.00 M⁻¹ = 0.0350 M
3. 3. Dilution factor = 50.0 mL ÷ 10.0 mL = 5.00
4. 4. Original concentration = 0.0350 M × 5.00 = 0.175 M

0.00700 M: Divides by the dilution factor instead of multiplying. The original sample is more concentrated than the diluted one.

0.0350 M: This is the concentration of the diluted solution read from the line; it does not correct for the 5-fold dilution.

0.175 M: Correct. Diluted: c = A ÷ slope = 0.280 ÷ 8.00 ${\mathrm{M}}^{-1}$ = 0.0350 M. Original: 0.0350 M × (50.0 mL ÷ 10.0 mL) = 0.175 M.

2.24 M: Multiplies the absorbance by the slope instead of dividing (inverts the calibration equation).

Read the measured signal back through the calibration line, then undo any dilution with M_original = M_diluted × (V_final ÷ V_aliquot).

Answer: C

Diagram A: Diagram A shows intact K₂SO₄ units with no separate ions. A soluble ionic compound dissociates into its ions.

Diagram B: Diagram B shows K⁺ and SO₄²⁻ in a 1 : 1 ratio (3 of each). The formula requires two K⁺ for every sulfate, and 1 : 1 would not balance the charges.

Diagram C: Correct. Diagram C shows 4 K⁺ ions and 2 intact SO₄²⁻ ions, the 2 : 1 ratio of the formula. Each sulfate stays together as one polyatomic ion, held by covalent S–O bonds.

Diagram D: Diagram D shows separate K, S and O atoms. The covalent bonds inside the sulfate ion do not break when the salt dissolves.

A strong electrolyte dissolves into its ions in the formula ratio, and polyatomic ions stay intact.