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(H2​O)m(H2​O)÷ M(X)× 2/ν× M(H2​O)

20 g of H2 and 32 g of O2 react according to 2H2 + O2 → 2H2O. What mass m of H2O can form?

M(H2) = 2.016M(O2) = 32M(H2O) = 18.02 g/mol
m = g
2.StoichiometryHard
m(Al)n(Al)n(AlCl3​)m(AlCl3​)÷ M(Al)× 2/2× M(AlCl3​)

In the reaction 2Al + 3Cl2 → 2AlCl3, 8 g of Al reacts completely (the other reactants are in excess) and 35.98 g of AlCl3 is collected. Find the percent yield Y.

M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 35.98 g
Y = %
3.Molarity & dilutionMedium
6 mol/LV1​0.4 mol/L1000 mL

You need 1000 mL of 0.4 mol/L solution and have a 6 mol/L stock solution. How much water Vwater do you add to the stock solution?

c1 = 6 mol/Lc2 = 0.4 mol/LV2 = 1000 mL
Vwater = mL
4.Molarity & dilutionHard
0.4 mol/L200 mL1.2 mol/L400 mLc = ?600 mL

200 mL of a 0.4 mol/L solution is mixed with 400 mL of a 1.2 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 0.4 mol/LV1 = 200 mLc2 = 1.2 mol/LV2 = 400 mL
c = mol/L
5.Ideal gas lawMedium
140 g of CO₂V = 20 LT = 20 °CP = ?

A tank of 20 L contains 140 g of CO₂ at 20 °C. Treat it as an ideal gas and find the pressure P.

m = 140 gM = 44.01 g/molV = 20 LT = 20 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
Ar, M = 39.95 g/molP = 900 kPaT = 10 °Cρ = ?

Find the density ρ of Ar at 900 kPa and 10 °C, treating it as an ideal gas.

M = 39.95 g/molP = 900 kPaT = 10 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.Calorimetry & enthalpyMedium
Tf​ = ?Copperm = 200 gT = 130 °Cc = 0.385 J/(g·°C)Waterm = 300 gT = 18 °Cc = 4.18 J/(g·°C)

A 200 g block of copper (c = 0.385 J/(g·°C)) at 130 °C is dropped into 300 g of water (c = 4.18 J/(g·°C)) at 18 °C in an insulated cup. Find the final temperature Tf.

mmetal = 200 gcmetal = 0.385 J/(g·°C)Tmetal = 130 °Cmwater = 300 gTwater = 18 °Ccwater = 4.18 J/(g·°C)
Tf = °C
8.Calorimetry & enthalpyHard
H2C2​H6​ + 7O₂reactants4CO₂ + 6H₂O(l)productsΔH° = ?

Use the standard enthalpies of formation to find ΔH° for the reaction 2C2H6 + 7O₂ → 4CO₂ + 6H₂O(l).

ΔH°f(C2H6) = −84.7 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​HII0.250.250C−x−x+2xE0.25 − x0.25 − x2x

0.25 mol/L of H₂ and 0.25 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 16. Find the equilibrium concentration [H2].

[H₂]₀ = [I₂]₀ = 0.25 mol/LKc = 16
[H2] = mol/L
10.Chemical equilibriumHard
PCl5​PCl3​Cl2​I0.200C−x+x+xE0.2 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.02 at the reaction temperature. A container starts with 0.2 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].

[PCl₅]₀ = 0.2 mol/LKc = 0.02
[PCl5] = mol/L
11.pH & buffersMedium
02468101214acidicneutralbasicpH = ?

Find the pH pH of a 0.2 mol/L solution of formic acid (Ka = 0.00018).

c = 0.2 mol/LKa = 0.00018
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

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

[HA] = 0.2[A⁻] = 0.2Ka = 0.00014
pH =
13.Reaction kineticsMedium
t (min)X (%)t = ?20%

A zero-order reaction has k = 0.005 mol/(L·min) and starts at 1 mol/L. How long does it take to reach 20% conversion? Find t.

k = 0.005 mol/(L·min)CA0 = 1 mol/LX = 20 %
t = min
14.Reaction kineticsHard
1/Tln kT = 25 °Cln 0.005ln A = ?Ea​ = 100 kJ/mol

A first-order reaction has a rate constant of 0.005 s⁻¹ at 25 °C and an activation energy of 100 kJ/mol. Find the pre-exponential factor A in the Arrhenius equation.

k = 0.005 s⁻¹T = 25 °CEa = 100 kJ/molR = 8.314 J/(mol·K)
A = s⁻¹

3Solutions

1.Stoichiometry

Answer: m = 36.04 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(H2) = 20/2.016 = 9.921 mol
n(O2) = 32/32 = 1 mol

Limiting reactant

9.921/2 = 4.96, 1/1 = 1 → O2

Mass of product

n(H2O) = 1·2 = 2 mol
m = 2·18.02 = 36.04 g

Answer

m = 36.04 g
2.Stoichiometry

Answer: Y = 91.01 %

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(Al) = 8/26.98 = 0.2965 mol
m = 0.2965·2/2·133.33 = 39.53 g

Percent yield

Y = 35.98/39.53·100 = 91.01%

Answer

Y = 91.01%
3.Molarity & dilution

Answer: Vwater = 933.3 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 = 0.4·1000/6 = 66.67 mL

Water to add

Vwater = 1000 − 66.67 = 933.3 mL

Answer

Vwater = 933.3 mL
4.Molarity & dilution

Answer: c = 0.9333 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 = 0.4·200 + 1.2·400 = 560 mmol

Concentration of the mixture

c = 560/600 = 0.9333 mol/L

Answer

c = 0.9333 mol/L
5.Ideal gas law

Answer: P = 387.7 kPa

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

Moles

n = m/M = 140/44.01 = 3.181 mol
T = 20 + 273.15 = 293.1 K

Pressure

P = nRT/V = 3.181·8.314·293.1/0.02 = 387.7 kPa

Answer

P = 387.7 kPa
6.Ideal gas law

Answer: ρ = 15.27 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 10 + 273.15 = 283.1 K

Density

ρ = 900·39.95/(8.314·283.1) = 15.27 kg/m³

Answer

ρ = 15.27 kg/m³
7.Calorimetry & enthalpy

Answer: Tf = 24.48 °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

200·0.385·(130 − Tf) = 300·4.18·(Tf − 18)

Solve for the final temperature

Tf = (77·130 + 1254·18)/(77 + 1254) = 24.48 °C

Answer

Tf = 24.48 °C
8.Calorimetry & enthalpy

Answer: ΔH° = −3119 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° = [4·(−393.5) + 6·(−285.8)] − [2·(−84.7) + 7·0] = −3119 kJ

Answer

ΔH° = −3119 kJ

Negative: the reaction gives off heat (exothermic).

9.Chemical equilibrium

Answer: [H2] = 0.08333 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

16 = (2x)²/(0.25 − x)²

Take the square root

4 = 2x/(0.25 − x) → x = 0.25·4/(2 + 4) = 0.1667 mol/L

Equilibrium concentration

[H₂] = 0.25 − x = 0.08333 mol/L

Answer

[H2] = 0.08333 mol/L
10.Chemical equilibrium

Answer: [PCl5] = 0.146 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

0.02 = x²/(0.2 − x) → x² + 0.02·x − 0.004 = 0

Solve the quadratic

x = (−0.02 + √(0.02² + 4·0.004))/2 = 0.05403 mol/L

Equilibrium concentration

[PCl₅] = 0.2 − 0.05403 = 0.146 mol/L

Answer

[PCl5] = 0.146 mol/L
11.pH & buffers

Answer: pH = 2.228

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 = 0.005911 mol/L

pH

pH = −log(0.005911) = 2.228

Answer

pH = 2.228
12.pH & buffers

Answer: pH = 3.854

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(0.2/0.2) = 3.854

Answer

pH = 3.854
13.Reaction kinetics

Answer: t = 40 min

Integrated rate laws: zero order: C = C₀ − kt (straight line, t½ = C₀/(2k)); first order: ln(C₀/C) = kt (t½ = ln 2/k); second order: 1/C − 1/C₀ = kt (t½ = 1/(kC₀)). Conversion X means C = C₀(1 − X).

Amount converted

CA0·X = 1·0.2 = 0.2 mol/L

Time

t = CA0·X/k = 0.2/0.005 = 40 min

Answer

t = 40 min
14.Reaction kinetics

Answer: A = 1.656 × 10¹⁵ s⁻¹

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 temperature

T = 25 + 273.15 = 298.1 K

Solve Arrhenius for A

Ea/(RT) = 100000/(8.314·298.1) = 40.34
A = k·e40.34 = 0.005·3.313 × 10¹⁷ = 1.656 × 10¹⁵ s⁻¹

Answer

A = 1.656 × 10¹⁵ s⁻¹

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