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(Fe)m(Fe)÷ M(X)× 2/ν× M(Fe)

12 g of Fe2O3 and 40 g of CO react according to Fe2O3 + 3CO → 2Fe + 3CO2. What mass m of Fe can form?

M(Fe2O3) = 159.69M(CO) = 28.01M(Fe) = 55.85 g/mol
m = g
2.StoichiometryHard
m(C3​H8​)n(C3​H8​)n(CO2​)m(CO2​)÷ M(C3​H8​)× 3/1× M(CO2​)

In the reaction C3H8 + 5O2 → 3CO2 + 4H2O, 16 g of C3H8 reacts completely (the other reactants are in excess) and 43.59 g of CO2 is collected. Find the percent yield Y.

M(C3H8) = 44.1 g/molM(CO2) = 44.01 g/molactual: 43.59 g
Y = %
3.Molarity & dilutionMedium
2.5 mol/LV1​0.2 mol/L500 mL

You need 500 mL of 0.2 mol/L solution and have a 2.5 mol/L stock solution. How much water Vwater do you add to the stock solution?

c1 = 2.5 mol/Lc2 = 0.2 mol/LV2 = 500 mL
Vwater = mL
4.Molarity & dilutionHard
1.5 mol/L50 mL0.75 mol/L200 mLc = ?250 mL

50 mL of a 1.5 mol/L solution is mixed with 200 mL of a 0.75 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 1.5 mol/LV1 = 50 mLc2 = 0.75 mol/LV2 = 200 mL
c = mol/L
5.Ideal gas lawMedium
210 g of HeV = 30 LT = 70 °CP = ?

A tank of 30 L contains 210 g of He at 70 °C. Treat it as an ideal gas and find the pressure P.

m = 210 gM = 4.003 g/molV = 30 LT = 70 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
CO₂, M = 44.01 g/molP = 500 kPaT = 140 °Cρ = ?

Find the density ρ of CO₂ at 500 kPa and 140 °C, treating it as an ideal gas.

M = 44.01 g/molP = 500 kPaT = 140 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.Calorimetry & enthalpyMedium
Tf​ = ?Leadm = 250 gT = 130 °Cc = 0.129 J/(g·°C)Waterm = 250 gT = 16 °Cc = 4.18 J/(g·°C)

A 250 g block of lead (c = 0.129 J/(g·°C)) at 130 °C is dropped into 250 g of water (c = 4.18 J/(g·°C)) at 16 °C in an insulated cup. Find the final temperature Tf.

mmetal = 250 gcmetal = 0.129 J/(g·°C)Tmetal = 130 °Cmwater = 250 gTwater = 16 °Ccwater = 4.18 J/(g·°C)
Tf = °C
8.Calorimetry & enthalpyHard
HC6​H12​O6​ + 6O₂reactants6CO₂ + 6H₂O(l)productsΔH° = ?

Use the standard enthalpies of formation to find ΔH° for the reaction C6H12O6 + 6O₂ → 6CO₂ + 6H₂O(l).

ΔH°f(C6H12O6) = −1274 kJ/molΔH°f(CO₂) = −393.5 kJ/molΔH°f(H₂O, l) = −285.8 kJ/molΔH°f(O₂) = 0
ΔH° = kJ
9.Chemical equilibriumMedium
H2​I2​HII0.80.80C−x−x+2xE0.8 − x0.8 − x2x

0.8 mol/L of H₂ and 0.8 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 54. Find the equilibrium concentration [H2].

[H₂]₀ = [I₂]₀ = 0.8 mol/LKc = 54
[H2] = mol/L
10.Chemical equilibriumHard
PCl5​PCl3​Cl2​I0.500C−x+x+xE0.5 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.2 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 = 0.2
[Cl2] = mol/L
11.pH & buffersMedium
02468101214acidicneutralbasicpH = ?

Find the pH pH of a 0.2 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).

c = 0.2 mol/LKa = 6.3 × 10⁻⁵
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

A buffer contains 0.2 mol/L formic acid and 0.3 mol/L of its sodium salt. Ka = 0.00018. Find the pH pH of the buffer.

[HA] = 0.2[A⁻] = 0.3Ka = 0.00018
pH =
13.Reaction kineticsMedium
t (min)X (%)t = ?95%

A first-order reaction in a batch reactor has k = 0.3 min⁻¹. How long does it take to reach 95% conversion? Find t.

k = 0.3 min⁻¹X = 95 %
t = min
14.Reaction kineticsHard
t (min)X (%)45 minX = ?

A first-order reaction has k = 0.005 min⁻¹ at 20 °C and an activation energy of 60 kJ/mol. The batch reactor is now run at 50 °C for 45 min. Find the conversion X reached.

k1 = 0.005 min⁻¹T1 = 20 °CEa = 60 kJ/molT2 = 50 °Ct = 45 minR = 8.314 J/(mol·K)
X = %

3Solutions

1.Stoichiometry

Answer: m = 8.394 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(Fe2O3) = 12/159.69 = 0.07515 mol
n(CO) = 40/28.01 = 1.428 mol

Limiting reactant

0.07515/1 = 0.07515, 1.428/3 = 0.476 → Fe2O3

Mass of product

n(Fe) = 0.07515·2 = 0.1503 mol
m = 0.1503·55.85 = 8.394 g

Answer

m = 8.394 g
2.Stoichiometry

Answer: Y = 91 %

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(C3H8) = 16/44.1 = 0.3628 mol
m = 0.3628·3/1·44.01 = 47.9 g

Percent yield

Y = 43.59/47.9·100 = 91%

Answer

Y = 91%
3.Molarity & dilution

Answer: Vwater = 460 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 = 0.2·500/2.5 = 40 mL

Water to add

Vwater = 500 − 40 = 460 mL

Answer

Vwater = 460 mL
4.Molarity & dilution

Answer: c = 0.9 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 = 1.5·50 + 0.75·200 = 225 mmol

Concentration of the mixture

c = 225/250 = 0.9 mol/L

Answer

c = 0.9 mol/L
5.Ideal gas law

Answer: P = 4989 kPa

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

Moles

n = m/M = 210/4.003 = 52.46 mol
T = 70 + 273.15 = 343.1 K

Pressure

P = nRT/V = 52.46·8.314·343.1/0.03 = 4989 kPa

Answer

P = 4989 kPa
6.Ideal gas law

Answer: ρ = 6.406 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 140 + 273.15 = 413.1 K

Density

ρ = 500·44.01/(8.314·413.1) = 6.406 kg/m³

Answer

ρ = 6.406 kg/m³
7.Calorimetry & enthalpy

Answer: Tf = 19.41 °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

250·0.129·(130 − Tf) = 250·4.18·(Tf − 16)

Solve for the final temperature

Tf = (32.25·130 + 1045·16)/(32.25 + 1045) = 19.41 °C

Answer

Tf = 19.41 °C
8.Calorimetry & enthalpy

Answer: ΔH° = −2802 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° = [6·(−393.5) + 6·(−285.8)] − [1·(−1274) + 6·0] = −2802 kJ

Answer

ΔH° = −2802 kJ

Negative: the reaction gives off heat (exothermic).

9.Chemical equilibrium

Answer: [H2] = 0.1712 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

54 = (2x)²/(0.8 − x)²

Take the square root

7.348 = 2x/(0.8 − x) → x = 0.8·7.348/(2 + 7.348) = 0.6288 mol/L

Equilibrium concentration

[H₂] = 0.8 − x = 0.1712 mol/L

Answer

[H2] = 0.1712 mol/L
10.Chemical equilibrium

Answer: [Cl2] = 0.2317 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

0.2 = x²/(0.5 − x) → x² + 0.2·x − 0.1 = 0

Solve the quadratic

x = (−0.2 + √(0.2² + 4·0.1))/2 = 0.2317 mol/L

Equilibrium concentration

[Cl₂] = x = 0.2317 mol/L

Answer

[Cl2] = 0.2317 mol/L
11.pH & buffers

Answer: pH = 2.454

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/(0.2 − x) → x = 0.003518 mol/L

pH

pH = −log(0.003518) = 2.454

Answer

pH = 2.454
12.pH & buffers

Answer: pH = 3.921

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 = 3.745

Henderson–Hasselbalch

pH = 3.745 + log(0.3/0.2) = 3.921

Answer

pH = 3.921
13.Reaction kinetics

Answer: t = 9.986 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 = 0.05

Solve for the time

t = −ln(1 − X)/k = −ln(0.05)/0.3 = 9.986 min

Answer

t = 9.986 min
14.Reaction kinetics

Answer: X = 89.05 %

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).

Rate constant at the new temperature

T1 = 293.1 K, T2 = 323.1 K
k2 = 0.005·e−(60000/8.314)·(1/323.1 − 1/293.1) = 0.04915 min⁻¹

Conversion

X = 1 − e−0.04915·45 = 1 − 0.1095 = 89.05 %

Answer

X = 89.05 %

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