SCH4801 ASSIGNMENT 3 2024 QUESTIONS WITH SOLUTIONS 1. Chemical Reactions
Question: Balance the following chemical equation:
Fe+O2→Fe2O3\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3Fe+O2
→Fe2O3
Answer: The balanced equation is:
4Fe+3O2→2Fe2O34\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\
text{O}_34Fe+3O2 →2Fe2O3
2. Stoichiometry
Question: How many grams of water are produced when 5.0 grams of hydrogen gas react with excess oxygen?
Answer:
1.Write the balanced equation: 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \
rightarrow 2\text{H}_2\text{O}2H2 +O2→2H2O
2.Calculate the moles of hydrogen gas: Molar mass of H2=2.0 g/mol\text{Molar mass of } \
text{H}_2 = 2.0 \text{ g/mol}Molar mass of H2 =2.0 g/mol Moles of H2=5.0 g2.0 g/mol=2.5
moles\text{Moles of } \text{H}_2 = \frac{5.0 \text{ g}}{2.0 \text{ g/mol}} = 2.5 \
text{ moles}Moles of H2 =2.0 g/mol5.0 g=2.5 moles
3.According to the balanced equation, 2 moles of H2\text{H}_2H2 produce 2 moles of H2O\text{H}_2\text{O}H2 O. Therefore, 2.5 moles of H2\text{H}_2H2 will produce 2.5 moles of H2O\text{H}_2\text{O}H2 O.
4.Calculate the mass of water: Molar mass of H2O=18.0 g/mol\text{Molar mass of } \
text{H}_2\text{O} = 18.0 \text{ g/mol}Molar mass of H2 O=18.0 g/mol Mass of H2O=2.5 moles×18.0 g/mol=45.0 g\text{Mass of } \text{H}_2\text{O} = 2.5 \text{ moles} \times 18.0
\text{ g/mol} = 45.0 \text{ g}Mass of H2 O=2.5 moles×18.0 g/mol=45.0 g
3. Acids and Bases
Question: What is the pH of a solution with a hydrogen ion concentration of 1.0×10−41.0 \times 10^{-4}1.0×10−4 M?
Answer:
1.Use the formula for pH: pH=−log [H+]\text{pH} = -\log[\
text{H}^+]pH=−log[H+]
2.Substitute the given concentration: pH=−log (1.0×10−4)\text{pH} = -\
log(1.0 \times 10^{-4})pH=−log(1.0×10−4) pH=4\text{pH} = 4pH=4 4. Thermodynamics
Question: Calculate the change in enthalpy (ΔH\Delta HΔH) if 100 J of energy is absorbed by a
system at constant pressure and the work done by the system is 50 J.
Answer:
1.Use the formula for enthalpy change: ΔH=q−W\Delta H = q - WΔH=q−W
2.Substitute the given values: ΔH=100 J−50 J\Delta H = 100 \text{ J} - 50 \
text{ J}ΔH=100 J−50 J ΔH=50 J\Delta H = 50 \text{ J}ΔH=50 J
5. Equilibrium
Question: For the reaction A⇌B\text{A} \rightleftharpoons \text{B}A⇌B, if the equilibrium concentrations are [A]=0.2 M[\text{A}] = 0.2 \text{ M}[A]=0.2 M and [B]=0.8 M[\text{B}] = 0.8 \text{ M}[B]=0.8 M, what is the equilibrium constant KcK_cKc?
Answer:
1.Write the expression for the equilibrium constant: Kc=[B][A]K_c = \frac{[\text{B}]}{[\
text{A}]}Kc=[A][B]
2.Substitute the concentrations: Kc=0.8 M0.2 MK_c = \frac{0.8 \text{ M}}{0.2 \text{ M}}Kc
=0.2 M0.8 M Kc=4K_c = 4Kc=4
6. Chemical Bonding
Question: Describe the differences between ionic and covalent bonds.
Answer:
●Ionic Bonds: Formed when one atom donates electrons to another atom, creating ions. The electrostatic attraction between the positively charged ion (cation) and the negatively charged ion (anion) holds them together. Typically occurs between metals and nonmetals (e.g., NaCl).
●Covalent Bonds: Formed when two atoms share one or more pairs of electrons. This type of bonding usually occurs between nonmetals. The shared electrons allow each atom to attain a full outer shell (e.g., H O). ₂
7. Solution Chemistry
Question: Calculate the molarity of a solution prepared by dissolving 10 grams of NaCl in 500 mL of water.
Answer: 1.Calculate the molar mass of NaCl: Molar mass of NaCl=22.99 (Na)+35.45 (Cl)=58.44 g/mol\text{Molar mass of NaCl} = 22.99 \text{ (Na)} + 35.45 \text{ (Cl)} = 58.44 \text{ g/mol}Molar mass of NaCl=22.99 (Na)+35.45 (Cl)=58.44 g/mol
2.Calculate the number of moles of NaCl: Moles of NaCl=10 g58.44 g/mol≈0.171 moles\text{Moles of NaCl} = \frac{10 \text{ g}}{58.44 \
text{ g/mol}} \approx 0.171 \text{ moles}Moles of NaCl=58.44 g/mol10 g
≈0.171 moles
3.Convert the volume from mL to L: 500 mL=0.5 L500 \text{ mL} = 0.5 \text{ L}500 mL=0.5 L
4.Calculate the molarity: Molarity=Moles of soluteVolume of solution (L)\text{Molarity} = \
frac{\text{Moles of solute}}{\text{Volume of solution (L)}}Molarity=Volume of solution (L)Moles of solute Molarity=0.171 moles0.5 L=0.342 M\text{Molarity} = \frac{0.171 \
text{ moles}}{0.5 \text{ L}} = 0.342 \text{ M}Molarity=0.5 L0.171 moles =0.342 M
8. Kinetics
Question: The reaction A→BA \rightarrow BA→B follows first-order kinetics. If the rate
constant kkk is 0.05 s−10.05 \text{ s}^{-1}0.05 s−1, what is the half-life of the reaction?
Answer:
1.For a first-order reaction, the half-life (t1/2t_{1/2}t1/2 ) is given by: t1/2=ln (2)kt_{1/2} = \
frac{\ln(2)}{k}t1/2=kln(2)
2.Substitute the rate constant: t1/2=ln (2)0.05 s−1t_{1/2} = \frac{\ln(2)}
{0.05 \text{ s}^{-1}}t1/2 =0.05 s−1ln(2) t1/2≈0.6930.05t_{1/2} \approx \
frac{0.693}{0.05}t1/2 ≈0.050.693 t1/2≈13.86 st_{1/2} \approx 13.86 \
text{ s}t1/2≈13.86 s
9. Electrochemistry
Question: Determine the standard electrode potential for the cell reaction:
Cu2+(aq)+Zn(s)→Cu(s)+Zn2+(aq)\text{Cu}^{2+}(aq) + \text{Zn}(s) \rightarrow \
text{Cu}(s) + \text{Zn}^{2+}(aq)Cu2+(aq)+Zn(s)→Cu(s)+Zn2+(aq)
Given:
●ECu2+/Cu∘=+0.34 VE^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \
text{ V}ECu2+/Cu∘ =+0.34 V
●EZn2+/Zn∘=−0.76 VE^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \
text{ V}EZn2+/Zn∘ =−0.76 V
Answer:
1.Use the formula for the cell potential: Ecell∘=Ecathode∘−Eanode∘E^\circ_{\
text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}Ecell∘
=Ecathode∘−Eanode∘
2.In this reaction, Cu is the cathode (reduction) and Zn is the anode (oxidation):
Ecell∘=ECu2+/Cu∘−EZn2+/Zn∘E^\circ_{\text{cell}} = E^\circ_{\
Question: Balance the following chemical equation:
Fe+O2→Fe2O3\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3Fe+O2
→Fe2O3
Answer: The balanced equation is:
4Fe+3O2→2Fe2O34\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\
text{O}_34Fe+3O2 →2Fe2O3
2. Stoichiometry
Question: How many grams of water are produced when 5.0 grams of hydrogen gas react with excess oxygen?
Answer:
1.Write the balanced equation: 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \
rightarrow 2\text{H}_2\text{O}2H2 +O2→2H2O
2.Calculate the moles of hydrogen gas: Molar mass of H2=2.0 g/mol\text{Molar mass of } \
text{H}_2 = 2.0 \text{ g/mol}Molar mass of H2 =2.0 g/mol Moles of H2=5.0 g2.0 g/mol=2.5
moles\text{Moles of } \text{H}_2 = \frac{5.0 \text{ g}}{2.0 \text{ g/mol}} = 2.5 \
text{ moles}Moles of H2 =2.0 g/mol5.0 g=2.5 moles
3.According to the balanced equation, 2 moles of H2\text{H}_2H2 produce 2 moles of H2O\text{H}_2\text{O}H2 O. Therefore, 2.5 moles of H2\text{H}_2H2 will produce 2.5 moles of H2O\text{H}_2\text{O}H2 O.
4.Calculate the mass of water: Molar mass of H2O=18.0 g/mol\text{Molar mass of } \
text{H}_2\text{O} = 18.0 \text{ g/mol}Molar mass of H2 O=18.0 g/mol Mass of H2O=2.5 moles×18.0 g/mol=45.0 g\text{Mass of } \text{H}_2\text{O} = 2.5 \text{ moles} \times 18.0
\text{ g/mol} = 45.0 \text{ g}Mass of H2 O=2.5 moles×18.0 g/mol=45.0 g
3. Acids and Bases
Question: What is the pH of a solution with a hydrogen ion concentration of 1.0×10−41.0 \times 10^{-4}1.0×10−4 M?
Answer:
1.Use the formula for pH: pH=−log [H+]\text{pH} = -\log[\
text{H}^+]pH=−log[H+]
2.Substitute the given concentration: pH=−log (1.0×10−4)\text{pH} = -\
log(1.0 \times 10^{-4})pH=−log(1.0×10−4) pH=4\text{pH} = 4pH=4 4. Thermodynamics
Question: Calculate the change in enthalpy (ΔH\Delta HΔH) if 100 J of energy is absorbed by a
system at constant pressure and the work done by the system is 50 J.
Answer:
1.Use the formula for enthalpy change: ΔH=q−W\Delta H = q - WΔH=q−W
2.Substitute the given values: ΔH=100 J−50 J\Delta H = 100 \text{ J} - 50 \
text{ J}ΔH=100 J−50 J ΔH=50 J\Delta H = 50 \text{ J}ΔH=50 J
5. Equilibrium
Question: For the reaction A⇌B\text{A} \rightleftharpoons \text{B}A⇌B, if the equilibrium concentrations are [A]=0.2 M[\text{A}] = 0.2 \text{ M}[A]=0.2 M and [B]=0.8 M[\text{B}] = 0.8 \text{ M}[B]=0.8 M, what is the equilibrium constant KcK_cKc?
Answer:
1.Write the expression for the equilibrium constant: Kc=[B][A]K_c = \frac{[\text{B}]}{[\
text{A}]}Kc=[A][B]
2.Substitute the concentrations: Kc=0.8 M0.2 MK_c = \frac{0.8 \text{ M}}{0.2 \text{ M}}Kc
=0.2 M0.8 M Kc=4K_c = 4Kc=4
6. Chemical Bonding
Question: Describe the differences between ionic and covalent bonds.
Answer:
●Ionic Bonds: Formed when one atom donates electrons to another atom, creating ions. The electrostatic attraction between the positively charged ion (cation) and the negatively charged ion (anion) holds them together. Typically occurs between metals and nonmetals (e.g., NaCl).
●Covalent Bonds: Formed when two atoms share one or more pairs of electrons. This type of bonding usually occurs between nonmetals. The shared electrons allow each atom to attain a full outer shell (e.g., H O). ₂
7. Solution Chemistry
Question: Calculate the molarity of a solution prepared by dissolving 10 grams of NaCl in 500 mL of water.
Answer: 1.Calculate the molar mass of NaCl: Molar mass of NaCl=22.99 (Na)+35.45 (Cl)=58.44 g/mol\text{Molar mass of NaCl} = 22.99 \text{ (Na)} + 35.45 \text{ (Cl)} = 58.44 \text{ g/mol}Molar mass of NaCl=22.99 (Na)+35.45 (Cl)=58.44 g/mol
2.Calculate the number of moles of NaCl: Moles of NaCl=10 g58.44 g/mol≈0.171 moles\text{Moles of NaCl} = \frac{10 \text{ g}}{58.44 \
text{ g/mol}} \approx 0.171 \text{ moles}Moles of NaCl=58.44 g/mol10 g
≈0.171 moles
3.Convert the volume from mL to L: 500 mL=0.5 L500 \text{ mL} = 0.5 \text{ L}500 mL=0.5 L
4.Calculate the molarity: Molarity=Moles of soluteVolume of solution (L)\text{Molarity} = \
frac{\text{Moles of solute}}{\text{Volume of solution (L)}}Molarity=Volume of solution (L)Moles of solute Molarity=0.171 moles0.5 L=0.342 M\text{Molarity} = \frac{0.171 \
text{ moles}}{0.5 \text{ L}} = 0.342 \text{ M}Molarity=0.5 L0.171 moles =0.342 M
8. Kinetics
Question: The reaction A→BA \rightarrow BA→B follows first-order kinetics. If the rate
constant kkk is 0.05 s−10.05 \text{ s}^{-1}0.05 s−1, what is the half-life of the reaction?
Answer:
1.For a first-order reaction, the half-life (t1/2t_{1/2}t1/2 ) is given by: t1/2=ln (2)kt_{1/2} = \
frac{\ln(2)}{k}t1/2=kln(2)
2.Substitute the rate constant: t1/2=ln (2)0.05 s−1t_{1/2} = \frac{\ln(2)}
{0.05 \text{ s}^{-1}}t1/2 =0.05 s−1ln(2) t1/2≈0.6930.05t_{1/2} \approx \
frac{0.693}{0.05}t1/2 ≈0.050.693 t1/2≈13.86 st_{1/2} \approx 13.86 \
text{ s}t1/2≈13.86 s
9. Electrochemistry
Question: Determine the standard electrode potential for the cell reaction:
Cu2+(aq)+Zn(s)→Cu(s)+Zn2+(aq)\text{Cu}^{2+}(aq) + \text{Zn}(s) \rightarrow \
text{Cu}(s) + \text{Zn}^{2+}(aq)Cu2+(aq)+Zn(s)→Cu(s)+Zn2+(aq)
Given:
●ECu2+/Cu∘=+0.34 VE^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \
text{ V}ECu2+/Cu∘ =+0.34 V
●EZn2+/Zn∘=−0.76 VE^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \
text{ V}EZn2+/Zn∘ =−0.76 V
Answer:
1.Use the formula for the cell potential: Ecell∘=Ecathode∘−Eanode∘E^\circ_{\
text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}Ecell∘
=Ecathode∘−Eanode∘
2.In this reaction, Cu is the cathode (reduction) and Zn is the anode (oxidation):
Ecell∘=ECu2+/Cu∘−EZn2+/Zn∘E^\circ_{\text{cell}} = E^\circ_{\
