Equilibrium
Define dynamic equilibrium
A state in a closed system where the rate of the forward reaction equals the rate of the reverse reaction, so concentrations remain constant.
Define a Brønsted-Lowry acid.
A proton (H⁺) donor.
What is oxidation?
Loss of electrons.
At which electrode does oxidation occur?
The anode.
What apparatus accurately delivers a titre?
A burette.
State the effect of increasing pressure on: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
The equilibrium shifts right (towards NH₃) because the product side has fewer moles of gas (2 vs 4).
A solution has pH = 3. Calculate [H⁺].
[H+]=10−pH --> [H+]=10−3 --> [H+]=1.0×10−3 mol L−1
Determine the oxidation number of sulfur in SO₄²⁻.
S+4(−2) =−2
S−8=−2
S=+6
State the direction of electron flow in a Zn/Cu galvanic cell.
From zinc electrode to copper electrode.
Calculate moles of HCl in 25.0 mL of 0.100 M HCl.
n=CV =(0.100)(0.0250) =2.50×10−3mol
Predict and explain the equilibrium shift when temperature is increased for an exothermic reaction.
The equilibrium shifts left (towards reactants). Heat is a product in an exothermic reaction, so adding heat causes the system to oppose the change by consuming heat.
Why does 0.10 M CH₃COOH have a higher pH than 0.10 M HCl?
HCl completely dissociates (strong acid), whereas CH₃COOH only partially dissociates (weak acid), producing fewer H⁺ ions.
Identify the oxidising and reducing agents.
Zn+Cu2+→Zn2++Cu
Calculate E°cell for: Cu²⁺/Cu = +0.34 V; Zn²⁺/Zn = −0.76 V
Ecell∘=Ecathode∘−Eanode∘ =0.34−(−0.76) =1.10V
Determine concentration of NaOH if 25.0 mL requires 20.0 mL of 0.150 M HCl.
Moles HCl: 0.150×0.0200=0.00300mol
1:1 ratio: n(NaOH)=0.00300mol
Concentration: C= 0.00300/0.0250 = 0.120M
The equilibrium 2NO₂(g) ⇌ N₂O₄(g) becomes darker when heated. Explain using Le Châtelier's Principle.
NO₂ is brown and N₂O₄ is colourless. Heating shifts the equilibrium toward the endothermic direction, producing more NO₂, so the mixture becomes darker.
Explain how a CH₃COOH/CH₃COO⁻ buffer resists pH change when acid is added.
Added H⁺ ions react with CH₃COO⁻ ions:
CH3COO−+H+→CH3COOH
The added hydrogen ions are removed, so pH changes only slightly.
Balance in acidic solution: MnO4−+Fe2+→Mn2++Fe3+
Half equations
MnO4−+ 8H++5e−→Mn2++4H2O
Fe2+→Fe3++e−
Multiply iron half-equation by 5 and combine:
MnO4−+8H++5Fe2+→Mn2++4H2O+5Fe3+
Molten sodium chloride is electrolysed using inert graphite electrodes. Write half-equations occurring at the cathode and the anode.
Cathode (reduction):Na++e−→Na(l)
Anode (oxidation): 2Cl−→Cl2(g)+2e−
Outline calculations to determine % NaHCO₃ in an antacid.
N₂O₄(g) ⇌ 2NO₂(g), ΔH > 0. Pressure decreases and temperature increases simultaneously.
A 25.0 mL sample of a CH3COOH requires 18.40 mL of 0.120 M NaOH. Calculate the [acid].
1. Calculate moles NaOH:
n=CV = (0.120) (0.01840) =2.21×10−3
2. CH3COOH reacts 1:1 n(acid)=2.21×10−3
=0.0250/2.21×10−3 =0.0883 M
Conclusion: The acid is weak.
0.0883M
Copper placed in AgNO₃ solution. Ag++e−→Ag E∘=+0.80V Cu2++2e−→Cu E∘=+0.34V
Ecell∘=0.80−0.34 =+0.46V Positive E° means spontaneous.
Half-equations:
Oxidation: Cu → Cu2++2e−
Reduction: 2Ag++2e−→2Ag
Cu+2Ag+→ Cu2++2Ag
Calculate the standard potential Mg/Mg²⁺ and Ag⁺/Ag galvanic cell. Ag++e−→AgE∘=+0.80V; Mg2++2e−→MgE∘=−2.37V
Ecell∘=0.80−(−2.37) =3.17V
Tablet mass = 0.465 g ; Solution volume = 100.0 mL Aliquot = 20.0 mL ; HCl = 0.168 M; Titre = 3.85 mL
n(HCl)=CV =(0.168)(0.00385) = 6.47×10−4mol
mole ratio 1:1 NaHCO3+HCl→NaCl+H2O+CO2
n(NaHCO3)=6.47×10−4mol
Scale to total solution:
6.47×10−4mol x (100.0/20.0) =3.24×10−3mol
m=n x M = (3.24×10−3mol)(84.01) = 0.272g
= 0.465 / 0.272g = 58.5%