General Chemistry: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

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1Formulas and methods

Stoichiometry

Moles
n = m/M
Concentration
c = n/V
Mole ratio
use the coefficients of the balanced equation
Limiting reactant
the one that runs out first
Percent yield
actual / theoretical × 100%

Method

  1. Write and balance the chemical equation.
  2. Convert the given mass to moles: n = m/M.
  3. Use the coefficients as the mole ratio to find the moles of the substance you want.
  4. Convert back to mass with m = n·M. With two reactants, find the limiting reactant first.

Molarity & dilution

Molarity
c = n/V (mol/L)
Moles
n = m/M
Dilution
c1V1 = c2V2
Mixing
c = (c1V1 + c2V2)/(V1 + V2)

Method

  1. Convert mass to moles with n = m/M.
  2. Convert the volume to litres.
  3. Molarity c = n/V in mol/L.
  4. For a dilution, the moles stay the same: c₁V₁ = c₂V₂.

Ideal gas law

Ideal gas law
pV = nRT
Gas constant
R = 8.314 J/(mol·K)
Combined gas law
p1V1/T1 = p2V2/T2
Moles
n = m/M
Temperature
T[K] = T[°C] + 273.15

Calorimetry & enthalpy

Heat
q = m·c·ΔT
Water
c = 4.18 J/(g·°C)
Calorimeter
heat lost = heat gained
Enthalpy of reaction
ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants)

Chemical equilibrium

Equilibrium constant
Kc = [C]c[D]d/([A]a[B]b)
ICE table
Initial, Change (×coefficient), Equilibrium
Quadratic
x = (−b + √(b² − 4ac))/(2a)

Method

  1. Write Kc with products over reactants, each raised to the power of its coefficient.
  2. Set up an ICE table: Initial, Change (in steps of x times the coefficients), Equilibrium.
  3. Put the equilibrium expressions into Kc and solve for x.
  4. Check whether x is small enough for an approximation; otherwise solve the quadratic.

pH & buffers

pH
pH = −log[H⁺]
pOH
pOH = −log[OH⁻], pH + pOH = 14
Weak acid
[H⁺] ≈ √(Ka·c)
Buffer
pH = pKa + log([A⁻]/[HA])
Dilution
c1V1 = c2V2

Method

  1. Decide whether the acid or base is strong (fully dissociated) or weak.
  2. Strong acid: [H⁺] equals its concentration. Strong base: find [OH⁻] first.
  3. Weak acid: solve Ka = x²/(c − x) for x = [H⁺].
  4. pH = −log[H⁺], and pH + pOH = 14 at 25 °C.

Reaction kinetics

First order
ln(CA/CA0) = −kt
Half-life (first order)
t½ = ln 2/k
Second order
1/CA = 1/CA0 + kt
Arrhenius
k = A·e−Ea/(RT)
Two temperatures
ln(k2/k1) = (Ea/R)(1/T1 − 1/T2)

2Practice problems

1.StoichiometryMedium
m(X)n(X)n(NaCl)m(NaCl)÷ M(X)× 2/ν× M(NaCl)

16 g of Na and 40 g of Cl2 react according to 2Na + Cl2 → 2NaCl. What mass m of NaCl can form?

M(Na) = 22.99M(Cl2) = 70.9M(NaCl) = 58.44 g/mol
m = g
2.StoichiometryHard
m(N2​)n(N2​)n(NH3​)m(NH3​)÷ M(N2​)× 2/1× M(NH3​)

In the reaction N2 + 3H2 → 2NH3, 12 g of N2 reacts completely (the other reactants are in excess) and 8.02 g of NH3 is collected. Find the percent yield Y.

M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 8.02 g
Y = %
3.Molarity & dilutionMedium
2 mol/LV1​ = ?0.25 mol/L200 mL

You need 200 mL of 0.25 mol/L solution and have a 2 mol/L stock solution. What volume V1 of stock solution do you need?

c1 = 2 mol/Lc2 = 0.25 mol/LV2 = 200 mL
V1 = mL
4.Molarity & dilutionHard
0.8 mol/L50 mL0.6 mol/L400 mLc = ?450 mL

50 mL of a 0.8 mol/L solution is mixed with 400 mL of a 0.6 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 0.8 mol/LV1 = 50 mLc2 = 0.6 mol/LV2 = 400 mL
c = mol/L
5.Ideal gas lawMedium
470 g of ArV = 30 LT = 85 °CP = ?

A tank of 30 L contains 470 g of Ar at 85 °C. Treat it as an ideal gas and find the pressure P.

m = 470 gM = 39.95 g/molV = 30 LT = 85 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of He goes from 250 kPa, 2 L, 10 °C to 600 kPa and 60 °C. Find the new volume V2.

P1 = 250 kPaV1 = 2 LT1 = 10 °CP2 = 600 kPaT2 = 60 °C
V2 = L
7.Calorimetry & enthalpyMedium
Tf​ = ?Aluminiumm = 100 gT = 160 °Cc = 0.897 J/(g·°C)Waterm = 200 gT = 20 °Cc = 4.18 J/(g·°C)

A 100 g block of aluminium (c = 0.897 J/(g·°C)) at 160 °C is dropped into 200 g of water (c = 4.18 J/(g·°C)) at 20 °C in an insulated cup. Find the final temperature Tf.

mmetal = 100 gcmetal = 0.897 J/(g·°C)Tmetal = 160 °Cmwater = 200 gTwater = 20 °Ccwater = 4.18 J/(g·°C)
Tf = °C
8.Calorimetry & enthalpyHard
H250 g C2​H5​OHC2​H5​OH + 3O₂reactants2CO₂ + 3H₂O(l)productsq = ?

Use the standard enthalpies of formation to find the heat q released when 250 g of C2H5OH (M = 46.07 g/mol) burns completely: C2H5OH + 3O₂ → 2CO₂ + 3H₂O(l). Give q as a positive number.

ΔH°f(C2H5OH) = −277.7 kJ/molΔH°f(CO₂) = −393.5 kJ/molΔH°f(H₂O, l) = −285.8 kJ/molΔH°f(O₂) = 0
q = kJ
9.Chemical equilibriumMedium
H2​I2​HII0.40.40C−x−x+2xE0.4 − x0.4 − x2x

0.4 mol/L of H₂ and 0.4 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 25. Find the equilibrium concentration [HI].

[H₂]₀ = [I₂]₀ = 0.4 mol/LKc = 25
[HI] = mol/L
10.Chemical equilibriumHard
PCl5​PCl3​Cl2​I0.500C−x+x+xE0.5 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.08 at the reaction temperature. A container starts with 0.5 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].

[PCl₅]₀ = 0.5 mol/LKc = 0.08
[Cl2] = mol/L
11.pH & buffersMedium
02468101214acidicneutralbasicpH = ?

Find the pH pH of a 0.25 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).

c = 0.25 mol/LKa = 6.3 × 10⁻⁵
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

A buffer contains 0.2 mol/L lactic acid and 0.05 mol/L of its sodium salt. Ka = 0.00014. Find the pH pH of the buffer.

[HA] = 0.2[A⁻] = 0.05Ka = 0.00014
pH =
13.Reaction kineticsMedium
t (min)X (%)t = ?99%

A first-order reaction in a batch reactor has k = 0.01 min⁻¹. How long does it take to reach 99% conversion? Find t.

k = 0.01 min⁻¹X = 99 %
t = min
14.Reaction kineticsHard
Tk20 °C0.0135 °C0.0271

The rate constant of a reaction is 0.01 min⁻¹ at 20 °C and 0.0271 min⁻¹ at 35 °C. Find the activation energy Ea.

k1 = 0.01 min⁻¹T1 = 20 °Ck2 = 0.0271 min⁻¹T2 = 35 °CR = 8.314 J/(mol·K)
Ea = kJ/mol

3Solutions

1.Stoichiometry

Answer: m = 40.67 g

Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.

Moles of each reactant

n(Na) = 16/22.99 = 0.696 mol
n(Cl2) = 40/70.9 = 0.5642 mol

Limiting reactant

0.696/2 = 0.348, 0.5642/1 = 0.5642 → Na

Mass of product

n(NaCl) = 0.348·2 = 0.696 mol
m = 0.696·58.44 = 40.67 g

Answer

m = 40.67 g
2.Stoichiometry

Answer: Y = 54.98 %

Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.

Theoretical yield

n(N2) = 12/28.02 = 0.4283 mol
m = 0.4283·2/1·17.03 = 14.59 g

Percent yield

Y = 8.02/14.59·100 = 54.98%

Answer

Y = 54.98%
3.Molarity & dilution

Answer: V1 = 25 mL

Molarity: concentration c = n/V in mol/L, with n = m/M. Diluting adds solvent but not solute, so the number of moles stays the same: c₁·V₁ = c₂·V₂. When two solutions of the same solute are mixed, the moles add up and so do the volumes.

Volume of stock solution

V1 = c2·V2/c1 = 0.25·200/2 = 25 mL

Answer

V1 = 25 mL
4.Molarity & dilution

Answer: c = 0.6222 mol/L

Molarity: concentration c = n/V in mol/L, with n = m/M. Diluting adds solvent but not solute, so the number of moles stays the same: c₁·V₁ = c₂·V₂. When two solutions of the same solute are mixed, the moles add up and so do the volumes.

Total amount

n = 0.8·50 + 0.6·400 = 280 mmol

Concentration of the mixture

c = 280/450 = 0.6222 mol/L

Answer

c = 0.6222 mol/L
5.Ideal gas law

Answer: P = 1168 kPa

Ideal gas law: PV = nRT, with n = m/M.

Moles

n = m/M = 470/39.95 = 11.76 mol
T = 85 + 273.15 = 358.1 K

Pressure

P = nRT/V = 11.76·8.314·358.1/0.03 = 1168 kPa

Answer

P = 1168 kPa
6.Ideal gas law

Answer: V2 = 0.9805 L

Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.

Absolute temperatures

T1 = 283.1 K, T2 = 333.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 2·(250/600)·(333.1/283.1) = 0.9805 L

Answer

V2 = 0.9805 L
7.Calorimetry & enthalpy

Answer: Tf = 33.57 °C

Heat and enthalpy: heating a substance takes q = m·c·ΔT. In a calorimeter the heat lost by the hot object equals the heat gained by the cold one. The enthalpy change of a reaction follows from enthalpies of formation: ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants), each multiplied by its coefficient.

Heat balance

100·0.897·(160 − Tf) = 200·4.18·(Tf − 20)

Solve for the final temperature

Tf = (89.7·160 + 836·20)/(89.7 + 836) = 33.57 °C

Answer

Tf = 33.57 °C
8.Calorimetry & enthalpy

Answer: q = 7416 kJ

Heat and enthalpy: heating a substance takes q = m·c·ΔT. In a calorimeter the heat lost by the hot object equals the heat gained by the cold one. The enthalpy change of a reaction follows from enthalpies of formation: ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants), each multiplied by its coefficient.

Enthalpy of reaction

ΔH° = [2·(−393.5) + 3·(−285.8)] − [1·(−277.7) + 3·0] = −1367 kJ

Heat for the given mass

n = 250/46.07 = 5.427 mol
q = n·|ΔH°/1| = 5.427·1367 = 7416 kJ

Answer

q = 7416 kJ
9.Chemical equilibrium

Answer: [HI] = 0.5714 mol/L

Chemical equilibrium: for aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]c[D]d/([A]a[B]b), using concentrations at equilibrium. An ICE table (Initial, Change, Equilibrium) links the starting amounts to the equilibrium amounts through one unknown x, scaled by the coefficients.

Equilibrium expression

25 = (2x)²/(0.4 − x)²

Take the square root

5 = 2x/(0.4 − x) → x = 0.4·5/(2 + 5) = 0.2857 mol/L

Equilibrium concentration

[HI] = 2x = 0.5714 mol/L

Answer

[HI] = 0.5714 mol/L
10.Chemical equilibrium

Answer: [Cl2] = 0.164 mol/L

Chemical equilibrium: for aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]c[D]d/([A]a[B]b), using concentrations at equilibrium. An ICE table (Initial, Change, Equilibrium) links the starting amounts to the equilibrium amounts through one unknown x, scaled by the coefficients.

Equilibrium expression

0.08 = x²/(0.5 − x) → x² + 0.08·x − 0.04 = 0

Solve the quadratic

x = (−0.08 + √(0.08² + 4·0.04))/2 = 0.164 mol/L

Equilibrium concentration

[Cl₂] = x = 0.164 mol/L

Answer

[Cl2] = 0.164 mol/L
11.pH & buffers

Answer: pH = 2.405

pH: pH = −log₁₀[H₃O⁺] and pOH = −log₁₀[OH⁻], with pH + pOH = 14 at 25 °C. Strong acids and bases ionise completely. A weak acid only partly: Ka = x²/(c − x), where x = [H₃O⁺]. A buffer of a weak acid and its conjugate base follows Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]).

Equilibrium

HA ⇌ H⁺ + A⁻: x mol/L of acid ionises.

Ka = x²/(0.25 − x) → x = 0.003937 mol/L

pH

pH = −log(0.003937) = 2.405

Answer

pH = 2.405
12.pH & buffers

Answer: pH = 3.252

pH: pH = −log₁₀[H₃O⁺] and pOH = −log₁₀[OH⁻], with pH + pOH = 14 at 25 °C. Strong acids and bases ionise completely. A weak acid only partly: Ka = x²/(c − x), where x = [H₃O⁺]. A buffer of a weak acid and its conjugate base follows Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]).

pKa

pKa = 3.854

Henderson–Hasselbalch

pH = 3.854 + log(0.05/0.2) = 3.252

Answer

pH = 3.252
13.Reaction kinetics

Answer: t = 460.5 min

Reaction rates: for a first-order reaction −dC/dt = kC, so C = C₀·e−kt and the half-life ln 2/k does not depend on concentration. For a second-order reaction 1/C = 1/C₀ + kt. The rate constant rises with temperature following Arrhenius: k = A·e−Ea/(RT).

Conversion and concentration

C/C0 = 1 − X = 0.01

Solve for the time

t = −ln(1 − X)/k = −ln(0.01)/0.01 = 460.5 min

Answer

t = 460.5 min
14.Reaction kinetics

Answer: Ea = 49.92 kJ/mol

Reaction rates: for a first-order reaction −dC/dt = kC, so C = C₀·e−kt and the half-life ln 2/k does not depend on concentration. For a second-order reaction 1/C = 1/C₀ + kt. The rate constant rises with temperature following Arrhenius: k = A·e−Ea/(RT).

Absolute temperatures

T1 = 293.1 K, T2 = 308.1 K

Solve Arrhenius for Ea

Ea = −R·ln(k2/k1)/(1/T2 − 1/T1) = −8.314·0.9969/(-1.661e-4) = 49916 J/mol
Ea = 49.92 kJ/mol

Answer

Ea = 49.92 kJ/mol

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