Chemical Reactions — Diagnostic Tests

Unit Tests

UT-1: Chemical Bonding

Question:

(a) Describe the difference between ionic, covalent, and dative covalent bonds, giving an example of each.

(b) Draw the dot-and-cross diagram for the ammonium ion, $\text{NH}_4^+$, identifying the dative covalent bond.

(c) Explain the term “electronegativity” and describe the trend in electronegativity across Period 2.

(d) A molecule has a permanent dipole moment. Explain what this means and describe the type of intermolecular force that results between such molecules.

Solution:

(a)

• Ionic bond: electrostatic attraction between oppositely charged ions formed by electron transfer from a metal to a non-metal. Example: $\text{NaCl}$ ($\text{Na}^+$ and $\text{Cl}^-$).
• Covalent bond: shared pair of electrons between two atoms, with each atom contributing one electron. Example: $\text{H} - \text{H}$.
• Dative covalent (coordinate) bond: a shared pair of electrons in which both electrons come from the same atom. Example: the bond between $\text{NH}_3$ and $\text{H}^+$ to form $\text{NH}_4^+$, where the nitrogen lone pair provides both electrons.

(b) In $\text{NH}_4^+$, nitrogen (Group 5) has 5 outer electrons. Three of these form normal covalent bonds with three hydrogen atoms (one electron each from nitrogen and one from each hydrogen). The fourth hydrogen ($\text{H}^+$, which has no electrons) forms a dative covalent bond with nitrogen using a lone pair from nitrogen. The resulting ion has a tetrahedral arrangement with four N—H bonds and a $+1$ charge overall. The dative covalent bond is formed by the nitrogen atom donating both electrons to the shared pair.

(c) Electronegativity is the ability of an atom to attract the bonding electrons in a covalent bond towards itself. Across Period 2 (from lithium to fluorine), electronegativity increases. This is because the nuclear charge increases (more protons) while the atomic radius decreases (electrons pulled closer), so the ability to attract bonding electrons increases.

(d) A permanent dipole moment arises when there is an uneven distribution of electron density in a molecule due to differences in electronegativity between bonded atoms. One end of the bond carries a partial negative charge ($\delta-$) and the other carries a partial positive charge ($\delta+$). The resulting intermolecular force is called permanent dipole-dipole interaction: the positive end of one polar molecule is attracted to the negative end of a neighbouring polar molecule. These are stronger than London dispersion forces but weaker than hydrogen bonds.

---

UT-2: Enthalpy Changes and Hess’s Law

Question:

(a) Define the terms “standard enthalpy of combustion” and “standard enthalpy of formation.”

(b) Use Hess’s Law to calculate the standard enthalpy of combustion of propane ($\text{C}_3\text{H}_8$) given the following data:

• $\Delta H_f^\circ[\text{C}_3\text{H}_8(g)] = -104\,\text{kJ mol}^{-1}$
• $\Delta H_f^\circ[\text{CO}_2(g)] = -394\,\text{kJ mol}^{-1}$
• $\Delta H_f^\circ[\text{H}_2\text{O}(l)] = -286\,\text{kJ mol}^{-1}$

(c) Explain why the standard enthalpy of combustion of propane is an exothermic process in terms of bond breaking and bond making.

(d) A calorimetry experiment determines the enthalpy of combustion of ethanol. The student heats $200\,\text{g}$ of water from $22.0\,^\circ\text{C}$ to $54.5\,^\circ\text{C}$ by burning $0.95\,\text{g}$ of ethanol. Calculate the experimental enthalpy change and identify two sources of error.

Solution:

(a) Standard enthalpy of combustion ($\Delta H_c^\circ$): the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions (298 K and 100 kPa), with all reactants and products in their standard states.

Standard enthalpy of formation ($\Delta H_f^\circ$): the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions.

(b) The combustion of propane: $\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l)$

Using Hess’s Law with enthalpies of formation:

$\Delta H_c^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$

$\Delta H_c^\circ = [3 \times (-394) + 4 \times (-286)] - [-104 + 5 \times 0]$

$\Delta H_c^\circ = [-1182 + (-1144)] - [-104]$

$\Delta H_c^\circ = -2326 + 104 = -2222\,\text{kJ mol}^{-1}$

(c) During combustion, bonds in the reactants ($\text{C}-\text{H}$, $\text{C}-\text{C}$, and $\text{O}=\text{O}$) must be broken, which is endothermic (absorbs energy). New bonds are formed in the products ($\text{C}=\text{O}$ in $\text{CO}_2$ and $\text{O}-\text{H}$ in $\text{H}_2\text{O}$), which is exothermic (releases energy). The bonds formed ($\text{C}=\text{O}$ and $\text{O}-\text{H}$) are stronger (release more energy) than the bonds broken ($\text{C}-\text{H}$, $\text{C}-\text{C}$, $\text{O}=\text{O}$), so the net energy change is negative (exothermic). More energy is released by bond making than is absorbed by bond breaking.

(d) Energy transferred to water: $q = mc\Delta T = 200 \times 4.18 \times (54.5 - 22.0) = 200 \times 4.18 \times 32.5 = 27170\,\text{J} = 27.17\,\text{kJ}$

Moles of ethanol: $n = m / M = 0.95 / 46.0 = 0.02065\,\text{mol}$

$\Delta H_c = -q / n = -27.17 / 0.02065 = -1316\,\text{kJ mol}^{-1}$

Two sources of error:

1. Heat loss to the surroundings (air, calorimeter) means not all heat from combustion is transferred to the water, making the experimental value less negative than the literature value.
2. Incomplete combustion of ethanol produces carbon monoxide instead of carbon dioxide, releasing less energy per mole.

---

UT-3: Chemical Equilibrium

Question:

(a) State Le Chatelier’s principle and explain how it predicts the effect of increasing pressure on the equilibrium: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$.

(b) Explain what is meant by the equilibrium constant, $K_c$. For the reaction in part (a), write the expression for $K_c$.

(c) A mixture of nitrogen and hydrogen is placed in a sealed container at $450\,^\circ\text{C}$. At equilibrium, the concentrations are: $[\text{N}_2] = 1.2\,\text{mol L}^{-1}$, $[\text{H}_2] = 3.6\,\text{mol L}^{-1}$, $[\text{NH}_3] = 0.8\,\text{mol L}^{-1}$. Calculate $K_c$ and comment on whether the position of equilibrium lies towards reactants or products.

(d) Explain why a catalyst does not change the position of equilibrium but does increase the rate at which equilibrium is reached.

Solution:

(a) Le Chatelier’s principle: if a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the system will adjust to oppose the change and restore a new equilibrium.

For the reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$, there are 4 moles of gas on the left and 2 moles of gas on the right. Increasing pressure shifts the equilibrium in the direction that reduces the number of moles of gas, i.e., to the right, towards ammonia. The system opposes the increase in pressure by reducing the total number of gas molecules.

(b) The equilibrium constant, $K_c$, is the ratio of the concentrations of products to reactants at equilibrium, with each concentration raised to the power of its stoichiometric coefficient in the balanced equation. $K_c$ is only affected by temperature, not by changes in concentration or pressure.

$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$

(c)

$K_c = \frac{(0.8)^2}{(1.2)(3.6)^3} = \frac{0.64}{1.2 \times 46.656} = \frac{0.64}{55.987} = 0.0114$

Since $K_c$ is much less than 1, the equilibrium position lies towards the reactants (nitrogen and hydrogen). The small value indicates that at equilibrium, the concentrations of reactants are much greater than the concentration of products.

(d) A catalyst works by providing an alternative reaction pathway with a lower activation energy, increasing the rate of both the forward and reverse reactions equally. Because both rates increase by the same proportion, the ratio of forward to reverse rate (which determines $K_c$) remains unchanged. Therefore the equilibrium position is unchanged, but equilibrium is reached faster because both forward and reverse reactions proceed more quickly.

---

Integration Tests

IT-1: Bonding, Energetics, and Equilibrium Combined

Question:

(a) Nitrogen dioxide ($\text{NO}_2$) is a brown gas that dimerises to form dinitrogen tetroxide ($\text{N}_2\text{O}_4$), a colourless gas: $2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)$. The forward reaction is exothermic. Predict and explain the effect of decreasing temperature on the equilibrium position and the intensity of the brown colour observed.

(b) The $\text{N}-\text{O}$ bond in $\text{NO}_2$ has significant covalent character with a bond polarity. Explain the origin of this polarity and identify the type of intermolecular forces present between $\text{NO}_2$ molecules.

(c) Given that $\Delta H_f^\circ[\text{NO}_2(g)] = +33\,\text{kJ mol}^{-1}$ and $\Delta H_f^\circ[\text{N}_2\text{O}_4(g)] = +10\,\text{kJ mol}^{-1}$, calculate the standard enthalpy change for the dimerisation reaction.

(d) Explain how Le Chatelier’s principle and the calculated enthalpy change together help predict how this equilibrium responds to changes in conditions in an industrial context.

Solution:

(a) The forward reaction (dimerisation of $\text{NO}_2$ to $\text{N}_2\text{O}_4$) is exothermic. Decreasing temperature favours the exothermic direction (the system opposes the cooling by producing heat). Therefore the equilibrium shifts to the right, producing more $\text{N}_2\text{O}_4$ and consuming $\text{NO}_2$. Since $\text{NO}_2$ is brown and $\text{N}_2\text{O}_4$ is colourless, decreasing temperature causes the brown colour to fade as the concentration of $\text{NO}_2$ decreases.

(b) The polarity of the $\text{N}-\text{O}$ bond arises from the difference in electronegativity between nitrogen ($\chi = 3.0$) and oxygen ($\chi = 3.5$). Oxygen is more electronegative, so it attracts the bonding electrons more strongly, creating a partial negative charge ($\delta-$) on oxygen and a partial positive charge ($\delta+$) on nitrogen. The intermolecular forces between $\text{NO}_2$ molecules include London dispersion forces (present between all molecules) and permanent dipole-dipole interactions (due to the polar nature of the molecule).

(c) The reaction is: $2\text{NO}_2(g) \rightarrow \text{N}_2\text{O}_4(g)$

$\Delta H^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$

$\Delta H^\circ = (+10) - 2 \times (+33) = 10 - 66 = -56\,\text{kJ mol}^{-1}$

The dimerisation reaction is exothermic with $\Delta H^\circ = -56\,\text{kJ mol}^{-1}$.

(d) The calculated enthalpy change confirms the forward reaction is exothermic. In an industrial context, this means that operating at lower temperatures will increase the yield of $\text{N}_2\text{O}_4$ (the desired product of dimerisation), according to Le Chatelier’s principle. However, lower temperatures also slow the rate of reaction, so a compromise temperature must be found that gives an acceptable rate while maintaining a reasonable yield. High pressure would also favour the dimerisation since 2 moles of gas form 1 mole of gas, reducing the total number of gas molecules. Understanding both the energetics ($\Delta H$) and equilibrium principles allows chemists to choose optimal conditions.

---

IT-2: Applied Equilibrium and Energetics

Question:

(a) The Contact Process involves the reaction: $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$, $\Delta H = -197\,\text{kJ mol}^{-1}$. Explain the industrial conditions used (temperature, pressure, catalyst) with reference to equilibrium principles and kinetics.

(b) Vanadium(V) oxide ($\text{V}_2\text{O}_5$) is used as a catalyst in the Contact Process. Explain why using a catalyst is economically important even though it does not change the equilibrium position.

(c) Calculate the value of $K_c$ for the Contact Process at a given temperature if the equilibrium concentrations are: $[\text{SO}_2] = 0.40\,\text{mol L}^{-1}$, $[\text{O}_2] = 0.20\,\text{mol L}^{-1}$, $[\text{SO}_3] = 0.60\,\text{mol L}^{-1}$.

(d) A student argues that since the reaction is exothermic, the reaction should be carried out at the lowest possible temperature. Evaluate this argument.

Solution:

(a)

• Temperature: The forward reaction is exothermic, so lower temperatures favour the production of $\text{SO}_3$. However, very low temperatures give slow reaction rates. The compromise temperature used is approximately $400$—$450\,^\circ\text{C}$, which gives a reasonable rate while still favouring the products to some extent.
• Pressure: There are 3 moles of gas on the left and 2 moles on the right, so high pressure favours the forward reaction (fewer moles of gas). High pressures ($1$—$2\,\text{atm}$) are used, though extremely high pressures are not economically necessary because even at moderate pressures the yield is acceptable.
• Catalyst: Vanadium(V) oxide increases the rate at which equilibrium is reached without changing the equilibrium position, allowing lower temperatures to be used without unacceptably slow rates.

(b) Although the catalyst does not change the equilibrium position or $K_c$, it dramatically increases the rate at which equilibrium is reached. This means the process runs faster, producing more $\text{SO}_3$ per unit time, which increases throughput and profitability. Without a catalyst, the reaction would be so slow at the compromise temperature that it would not be economically viable. The catalyst allows the process to operate at a lower temperature (where equilibrium yields are still acceptable) while maintaining a fast rate.

(c)

$K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]} = \frac{(0.60)^2}{(0.40)^2(0.20)} = \frac{0.36}{0.032} = 11.25$

The value of $K_c = 11.25$, indicating that the equilibrium lies towards products ($\text{SO}_3$) at this temperature.

(d) While it is true that lower temperatures favour the exothermic forward reaction, the student’s argument is incomplete. At very low temperatures, the rate of reaction becomes impractically slow, even with a catalyst. The reaction would take so long that the process would not be economically viable. In practice, a compromise temperature must be chosen that balances a reasonable equilibrium yield with an acceptable reaction rate. The catalyst helps by allowing this compromise to be at a lower temperature than would otherwise be possible, but there is a practical lower limit.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

• Forgetting to include state symbols in equilibrium expressions and $K_c$ calculations.
• Misapplying Le Chatelier’s principle: the system opposes the change, it does not reverse it.
• Confusing the effect of a catalyst on equilibrium position (no effect) with its effect on rate (increases both forward and reverse equally).
• Errors in Hess’s Law calculations: mixing up the sign convention or forgetting to multiply by stoichiometric coefficients.
• Stating that $K_c$ changes with concentration or pressure — $K_c$ only changes with temperature.