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)× 1/ν× M(CO2​)

20 g of CH4 and 64 g of O2 react according to CH4 + 2O2 → CO2 + 2H2O. What mass m of CO2 can form?

M(CH4) = 16.04M(O2) = 32M(CO2) = 44.01 g/mol
m = g
2.StoichiometryHard
m(CaCO3​)n(CaCO3​)n(CO2​)m(CO2​)÷ M(CaCO3​)× 1/1× M(CO2​)

In the reaction CaCO3 → CaO + CO2, 24 g of CaCO3 reacts completely (the other reactants are in excess) and 8.23 g of CO2 is collected. Find the percent yield Y.

M(CaCO3) = 100.09 g/molM(CO2) = 44.01 g/molactual: 8.23 g
Y = %
3.Molarity & dilutionMedium
2.5 mol/LV1​ = ?0.3 mol/L100 mL

You need 100 mL of 0.3 mol/L solution and have a 2.5 mol/L stock solution. What volume V1 of stock solution do you need?

c1 = 2.5 mol/Lc2 = 0.3 mol/LV2 = 100 mL
V1 = mL
4.Molarity & dilutionHard
0.5 mol/L100 mL0.6 mol/L250 mLc = ?350 mL

100 mL of a 0.5 mol/L solution is mixed with 250 mL of a 0.6 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 0.5 mol/LV1 = 100 mLc2 = 0.6 mol/LV2 = 250 mL
c = mol/L
5.Ideal gas lawMedium
330 g of HeP = 100 kPaT = 80 °CV = ?

What volume V does 330 g of He occupy at 100 kPa and 80 °C? Treat it as an ideal gas.

m = 330 gM = 4.003 g/molP = 100 kPaT = 80 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
He, M = 4.003 g/molP = 700 kPaT = 200 °Cρ = ?

Find the density ρ of He at 700 kPa and 200 °C, treating it as an ideal gas.

M = 4.003 g/molP = 700 kPaT = 200 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.Calorimetry & enthalpyMedium
Tf​ = ?Aluminiumm = 50 gT = 150 °Cc = 0.897 J/(g·°C)Waterm = 400 gT = 16 °Cc = 4.18 J/(g·°C)

A 50 g block of aluminium (c = 0.897 J/(g·°C)) at 150 °C is dropped into 400 g of water (c = 4.18 J/(g·°C)) at 16 °C in an insulated cup. Find the final temperature Tf.

mmetal = 50 gcmetal = 0.897 J/(g·°C)Tmetal = 150 °Cmwater = 400 gTwater = 16 °Ccwater = 4.18 J/(g·°C)
Tf = °C
8.Calorimetry & enthalpyHard
H25 g C4​H10​2C4​H10​ + 13O₂reactants8CO₂ + 10H₂O(l)productsq = ?

Use the standard enthalpies of formation to find the heat q released when 25 g of C4H10 (M = 58.12 g/mol) burns completely: 2C4H10 + 13O₂ → 8CO₂ + 10H₂O(l). Give q as a positive number.

ΔH°f(C4H10) = −125.6 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​HII110C−x−x+2xE1 − x1 − x2x

1 mol/L of H₂ and 1 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 36. Find the equilibrium concentration [HI].

[H₂]₀ = [I₂]₀ = 1 mol/LKc = 36
[HI] = 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 [Cl2].

[PCl₅]₀ = 0.2 mol/LKc = 0.02
[Cl2] = mol/L
11.pH & buffersMedium
02468101214acidicneutralbasicpH = ?

Find the pH pH of a 0.25 mol/L solution of formic acid (Ka = 0.00018).

c = 0.25 mol/LKa = 0.00018
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

A 1.00 L buffer contains 0.3 mol hypochlorous acid and 0.1 mol of its conjugate base (Ka = 3 × 10⁻⁸). 0.025 mol of NaOH is added. Find the new pH pH.

n(HA) = 0.3 moln(A⁻) = 0.1 moln(OH⁻) = 0.025 molKa = 3 × 10⁻⁸
pH =
13.Reaction kineticsMedium
t (min)X (%)t = ?80%

A zero-order reaction has k = 0.025 mol/(L·min) and starts at 2 mol/L. How long does it take to reach 80% conversion? Find t.

k = 0.025 mol/(L·min)CA0 = 2 mol/LX = 80 %
t = min
14.Reaction kineticsHard
1/Tln kT = 25 °Cln 0.1ln A = ?Ea​ = 80 kJ/mol

A first-order reaction has a rate constant of 0.1 s⁻¹ at 25 °C and an activation energy of 80 kJ/mol. Find the pre-exponential factor A in the Arrhenius equation.

k = 0.1 s⁻¹T = 25 °CEa = 80 kJ/molR = 8.314 J/(mol·K)
A = s⁻¹

3Solutions

1.Stoichiometry

Answer: m = 44.01 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(CH4) = 20/16.04 = 1.247 mol
n(O2) = 64/32 = 2 mol

Limiting reactant

1.247/1 = 1.247, 2/2 = 1 → O2

Mass of product

n(CO2) = 1·1 = 1 mol
m = 1·44.01 = 44.01 g

Answer

m = 44.01 g
2.Stoichiometry

Answer: Y = 77.99 %

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) = 24/100.09 = 0.2398 mol
m = 0.2398·1/1·44.01 = 10.55 g

Percent yield

Y = 8.23/10.55·100 = 77.99%

Answer

Y = 77.99%
3.Molarity & dilution

Answer: V1 = 12 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 = 0.3·100/2.5 = 12 mL

Answer

V1 = 12 mL
4.Molarity & dilution

Answer: c = 0.5714 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 = 0.5·100 + 0.6·250 = 200 mmol

Concentration of the mixture

c = 200/350 = 0.5714 mol/L

Answer

c = 0.5714 mol/L
5.Ideal gas law

Answer: V = 2420 L

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

Moles

n = 330/4.003 = 82.44 mol
T = 80 + 273.15 = 353.1 K

Volume

V = nRT/P = 82.44·8.314·353.1/100 = 2420 L

Answer

V = 2420 L
6.Ideal gas law

Answer: ρ = 0.7123 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 200 + 273.15 = 473.1 K

Density

ρ = 700·4.003/(8.314·473.1) = 0.7123 kg/m³

Answer

ρ = 0.7123 kg/m³
7.Calorimetry & enthalpy

Answer: Tf = 19.5 °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

50·0.897·(150 − Tf) = 400·4.18·(Tf − 16)

Solve for the final temperature

Tf = (44.85·150 + 1672·16)/(44.85 + 1672) = 19.5 °C

Answer

Tf = 19.5 °C
8.Calorimetry & enthalpy

Answer: q = 1238 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° = [8·(−393.5) + 10·(−285.8)] − [2·(−125.6) + 13·0] = −5755 kJ

Heat for the given mass

n = 25/58.12 = 0.4301 mol
q = n·|ΔH°/2| = 0.4301·2877 = 1238 kJ

Answer

q = 1238 kJ
9.Chemical equilibrium

Answer: [HI] = 1.5 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

36 = (2x)²/(1 − x)²

Take the square root

6 = 2x/(1 − x) → x = 1·6/(2 + 6) = 0.75 mol/L

Equilibrium concentration

[HI] = 2x = 1.5 mol/L

Answer

[HI] = 1.5 mol/L
10.Chemical equilibrium

Answer: [Cl2] = 0.05403 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

[Cl₂] = x = 0.05403 mol/L

Answer

[Cl2] = 0.05403 mol/L
11.pH & buffers

Answer: pH = 2.179

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 = 0.006619 mol/L

pH

pH = −log(0.006619) = 2.179

Answer

pH = 2.179
12.pH & buffers

Answer: pH = 7.18

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]).

Neutralisation

n(HA) = 0.3 − 0.025 = 0.275 mol
n(A⁻) = 0.1 + 0.025 = 0.125 mol

Henderson–Hasselbalch

pH = 7.523 + log(0.125/0.275) = 7.18

Answer

pH = 7.18
13.Reaction kinetics

Answer: t = 64 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 = 2·0.8 = 1.6 mol/L

Time

t = CA0·X/k = 1.6/0.025 = 64 min

Answer

t = 64 min
14.Reaction kinetics

Answer: A = 1.038 × 10¹³ 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) = 80000/(8.314·298.1) = 32.27
A = k·e32.27 = 0.1·1.038 × 10¹⁴ = 1.038 × 10¹³ s⁻¹

Answer

A = 1.038 × 10¹³ s⁻¹

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