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(CO2​)m(CO2​)÷ M(X)× 3/ν× M(CO2​)

24 g of C3H8 and 64 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of CO2 can form?

M(C3H8) = 44.1M(O2) = 32M(CO2) = 44.01 g/mol
m = g
2.StoichiometryHard
m(CaCO3​)n(CaCO3​)n(CaO)m(CaO)÷ M(CaCO3​)× 1/1× M(CaO)

In the reaction CaCO3 → CaO + CO2, 8 g of CaCO3 reacts completely (the other reactants are in excess) and 3.5 g of CaO is collected. Find the percent yield Y.

M(CaCO3) = 100.09 g/molM(CaO) = 56.08 g/molactual: 3.5 g
Y = %
3.Molarity & dilutionMedium
5 mol/LV1​0.5 mol/L500 mL

You need 500 mL of 0.5 mol/L solution and have a 5 mol/L stock solution. How much water Vwater do you add to the stock solution?

c1 = 5 mol/Lc2 = 0.5 mol/LV2 = 500 mL
Vwater = mL
4.Molarity & dilutionHard
1.5 mol/L100 mL0.1 mol/L400 mLc = ?500 mL

100 mL of a 1.5 mol/L solution is mixed with 400 mL of a 0.1 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 1.5 mol/LV1 = 100 mLc2 = 0.1 mol/LV2 = 400 mL
c = mol/L
5.Ideal gas lawMedium
430 g of CH₄V = 50 LT = 70 °CP = ?

A tank of 50 L contains 430 g of CH₄ at 70 °C. Treat it as an ideal gas and find the pressure P.

m = 430 gM = 16.04 g/molV = 50 LT = 70 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of CO₂ goes from 200 kPa, 20 L, 30 °C to 400 kPa and 250 °C. Find the new volume V2.

P1 = 200 kPaV1 = 20 LT1 = 30 °CP2 = 400 kPaT2 = 250 °C
V2 = L
7.Calorimetry & enthalpyMedium
Tf​ = ?Aluminiumm = 150 gT = 170 °Cc = 0.897 J/(g·°C)Waterm = 250 gT = 16 °Cc = 4.18 J/(g·°C)

A 150 g block of aluminium (c = 0.897 J/(g·°C)) at 170 °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 = 150 gcmetal = 0.897 J/(g·°C)Tmetal = 170 °Cmwater = 250 gTwater = 16 °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​HII0.20.20C−x−x+2xE0.2 − x0.2 − x2x

0.2 mol/L of H₂ and 0.2 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 49. Find the equilibrium concentration [HI].

[H₂]₀ = [I₂]₀ = 0.2 mol/LKc = 49
[HI] = mol/L
10.Chemical equilibriumHard
PCl5​PCl3​Cl2​I100C−x+x+xE1 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.08 at the reaction temperature. A container starts with 1 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].

[PCl₅]₀ = 1 mol/LKc = 0.08
[Cl2] = mol/L
11.pH & buffersMedium
02468101214acidicneutralbasicpH = ?

Find the pH pH of a 0.05 mol/L solution of acetic acid (Ka = 1.8 × 10⁻⁵).

c = 0.05 mol/LKa = 1.8 × 10⁻⁵
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

A buffer contains 0.1 mol/L hydrofluoric acid and 0.5 mol/L of its sodium salt. Ka = 0.00068. Find the pH pH of the buffer.

[HA] = 0.1[A⁻] = 0.5Ka = 0.00068
pH =
13.Reaction kineticsMedium
t (min)C (mol/L)20 minC = ?

A reaction with rate law −rA = k·CA² has k = 0.5 L/(mol·min) and starts at 2 mol/L. Find the concentration CA after 20 min.

k = 0.5 L/(mol·min)CA0 = 2 mol/Lt = 20 min
CA = mol/L
14.Reaction kineticsHard
Tk20 °C0.0130 °C0.0295

The rate constant of a reaction is 0.01 min⁻¹ at 20 °C and 0.0295 min⁻¹ at 30 °C. Find the activation energy Ea.

k1 = 0.01 min⁻¹T1 = 20 °Ck2 = 0.0295 min⁻¹T2 = 30 °CR = 8.314 J/(mol·K)
Ea = kJ/mol

3Solutions

1.Stoichiometry

Answer: m = 52.81 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(C3H8) = 24/44.1 = 0.5442 mol
n(O2) = 64/32 = 2 mol

Limiting reactant

0.5442/1 = 0.5442, 2/5 = 0.4 → O2

Mass of product

n(CO2) = 0.4·3 = 1.2 mol
m = 1.2·44.01 = 52.81 g

Answer

m = 52.81 g
2.Stoichiometry

Answer: Y = 78.08 %

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(CaCO3) = 8/100.09 = 0.07993 mol
m = 0.07993·1/1·56.08 = 4.482 g

Percent yield

Y = 3.5/4.482·100 = 78.08%

Answer

Y = 78.08%
3.Molarity & dilution

Answer: Vwater = 450 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 = 0.5·500/5 = 50 mL

Water to add

Vwater = 500 − 50 = 450 mL

Answer

Vwater = 450 mL
4.Molarity & dilution

Answer: c = 0.38 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 = 1.5·100 + 0.1·400 = 190 mmol

Concentration of the mixture

c = 190/500 = 0.38 mol/L

Answer

c = 0.38 mol/L
5.Ideal gas law

Answer: P = 1530 kPa

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

Moles

n = m/M = 430/16.04 = 26.81 mol
T = 70 + 273.15 = 343.1 K

Pressure

P = nRT/V = 26.81·8.314·343.1/0.05 = 1530 kPa

Answer

P = 1530 kPa
6.Ideal gas law

Answer: V2 = 17.26 L

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

Absolute temperatures

T1 = 303.1 K, T2 = 523.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 20·(200/400)·(523.1/303.1) = 17.26 L

Answer

V2 = 17.26 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

150·0.897·(170 − Tf) = 250·4.18·(Tf − 16)

Solve for the final temperature

Tf = (134.6·170 + 1045·16)/(134.6 + 1045) = 33.57 °C

Answer

Tf = 33.57 °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: [HI] = 0.3111 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

49 = (2x)²/(0.2 − x)²

Take the square root

7 = 2x/(0.2 − x) → x = 0.2·7/(2 + 7) = 0.1556 mol/L

Equilibrium concentration

[HI] = 2x = 0.3111 mol/L

Answer

[HI] = 0.3111 mol/L
10.Chemical equilibrium

Answer: [Cl2] = 0.2457 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/(1 − x) → x² + 0.08·x − 0.08 = 0

Solve the quadratic

x = (−0.08 + √(0.08² + 4·0.08))/2 = 0.2457 mol/L

Equilibrium concentration

[Cl₂] = x = 0.2457 mol/L

Answer

[Cl2] = 0.2457 mol/L
11.pH & buffers

Answer: pH = 3.027

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/(0.05 − x) → x = 0.0009397 mol/L

pH

pH = −log(0.0009397) = 3.027

Answer

pH = 3.027
12.pH & buffers

Answer: pH = 3.866

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 = 3.167

Henderson–Hasselbalch

pH = 3.167 + log(0.5/0.1) = 3.866

Answer

pH = 3.866
13.Reaction kinetics

Answer: CA = 0.09524 mol/L

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).

Integrated second-order law

1/CA = 1/CA0 + kt = 0.5 + 0.5·20 = 10.5 L/mol
CA = 1/10.5 = 0.09524 mol/L

Answer

CA = 0.09524 mol/L
14.Reaction kinetics

Answer: Ea = 79.93 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 = 303.1 K

Solve Arrhenius for Ea

Ea = −R·ln(k2/k1)/(1/T2 − 1/T1) = −8.314·1.082/(-1.125e-4) = 79929 J/mol
Ea = 79.93 kJ/mol

Answer

Ea = 79.93 kJ/mol

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