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IB Chemistry – Reactivity 1: What Drives Reactions

Enthalpy, Hess’s law, Born-Haber cycles, entropy and Gibbs energy, electrochemistry.

12 questions  ·  37 marks  ·  answers included  ·  IB Chemistry SL & HL (2023 syllabus)
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This worksheet covers:

  • Questions on enthalpy changes
  • Hess’s law
  • Born-Haber cycles
  • Entropy and Gibbs energy
  • Electrochemistry

Use $R=8.31\,\text{J K}^{-1}\text{mol}^{-1}$, $F=96500\,\text{C mol}^{-1}$ and the specific heat capacity of water $c=4.18\,\text{J g}^{-1}\text{K}^{-1}$. A calculator is required throughout; Born-Haber and Gibbs energy calculations are HL extension content, marked accordingly.

Questions

Section A – Enthalpy and Hess’s Law

SL and HL
1

In a calorimetry experiment, burning a small sample of fuel raises the temperature of $100\,\text{g}$ of water by $6.5\degree\text{C}$. The mass of fuel burned corresponds to $0.0050\,\text{mol}$.(a) Calculate the heat energy transferred to the water, in kJ.(b) Calculate the molar enthalpy of combustion of the fuel, including its sign.

Show answer
(a) $q=mc\Delta T=100\times4.18\times6.5=2717\,\text{J}=2.72\,\text{kJ}$.(b) $\Delta H_c=-\dfrac{2.72}{0.0050}=-543\,\text{kJ mol}^{-1}$ (negative because combustion is exothermic).
[4]
2

State Hess’s law, and explain why it is a consequence of the first law of thermodynamics (conservation of energy).

Show answer
Hess’s law states that the total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same. This follows from conservation of energy: since enthalpy is a state function, the total energy change between two states cannot depend on the path, otherwise energy could be created or destroyed by choosing a different route.
[2]
3

Given the standard enthalpies of combustion $\Delta H_c^{\ominus}[\text{C(s)}]=-393.5\,\text{kJ mol}^{-1}$, $\Delta H_c^{\ominus}[\text{H}_2\text{(g)}]=-285.8\,\text{kJ mol}^{-1}$ and $\Delta H_c^{\ominus}[\text{CH}_4\text{(g)}]=-890.3\,\text{kJ mol}^{-1}$, use Hess’s law to calculate the standard enthalpy of formation of methane, $\Delta H_f^{\ominus}[\text{CH}_4\text{(g)}]$, for the reaction $\text{C(s)}+2\text{H}_2\text{(g)}\rightarrow\text{CH}_4\text{(g)}$.

Show answer
$\Delta H_f^{\ominus}=\Delta H_c^{\ominus}[\text{C}]+2\Delta H_c^{\ominus}[\text{H}_2]-\Delta H_c^{\ominus}[\text{CH}_4]=(-393.5)+2(-285.8)-(-890.3)=-74.8\,\text{kJ mol}^{-1}$.
[3]
4

Explain, in terms of bond breaking and bond making, why some reactions are exothermic and others are endothermic.

Show answer
Breaking bonds in the reactants requires an input of energy (endothermic), while forming bonds in the products releases energy (exothermic). If the energy released on forming the new bonds is greater than the energy required to break the old bonds, the overall reaction is exothermic ($\Delta H<0$); if less energy is released than absorbed, the reaction is endothermic ($\Delta H>0$).
[3]

Section B – Born-Haber Cycles, Entropy and Gibbs Energy

HL
5

Use the following data to construct a Born-Haber cycle for sodium chloride and calculate the lattice enthalpy of NaCl:Enthalpy of atomisation of Na(s): $+107\,\text{kJ mol}^{-1}$Enthalpy of atomisation of $\frac{1}{2}\text{Cl}_2$(g): $+122\,\text{kJ mol}^{-1}$First ionisation energy of Na(g): $+496\,\text{kJ mol}^{-1}$First electron affinity of Cl(g): $-349\,\text{kJ mol}^{-1}$Enthalpy of formation of NaCl(s): $-411\,\text{kJ mol}^{-1}$

Show answer
$\Delta H_f^{\ominus}=\Delta H_{\text{atomisation}}(\text{Na})+\Delta H_{\text{atomisation}}(\text{Cl})+IE_1(\text{Na})+EA_1(\text{Cl})+\Delta H_{\text{lattice}}$. Rearranging: $\Delta H_{\text{lattice}}=-411-(107+122+496-349)=-787\,\text{kJ mol}^{-1}$.
[4]
6

Define lattice enthalpy, and explain why lattice enthalpy values are always exothermic (negative).

Show answer
Lattice enthalpy is the enthalpy change when one mole of a solid ionic compound is formed from its gaseous ions under standard conditions. It is always exothermic because oppositely charged gaseous ions are attracted to each other by electrostatic forces, so bringing them together to form a stable ionic lattice releases energy.
[2]
7

For the reaction $\text{N}_2\text{(g)}+3\text{H}_2\text{(g)}\rightleftharpoons2\text{NH}_3\text{(g)}$, $\Delta H^{\ominus}=-92\,\text{kJ mol}^{-1}$ and $\Delta S^{\ominus}=-198\,\text{J K}^{-1}\text{mol}^{-1}$.(a) Calculate $\Delta G^{\ominus}$ at $298\,\text{K}$, and state whether the reaction is spontaneous at this temperature.(b) Calculate the temperature above which the reaction becomes non-spontaneous under standard conditions.

Show answer
(a) $\Delta G^{\ominus}=\Delta H^{\ominus}-T\Delta S^{\ominus}=-92-(298)(-0.198)=-92+59.0=-33.0\,\text{kJ mol}^{-1}$. Since $\Delta G^{\ominus}<0$, the reaction is spontaneous at $298\,\text{K}$.(b) The reaction becomes non-spontaneous when $\Delta G^{\ominus}=0$: $T=\dfrac{\Delta H^{\ominus}}{\Delta S^{\ominus}}=\dfrac{-92}{-0.198}=465\,\text{K}$.
[5]
8

Explain why $\Delta S^{\ominus}$ is negative for the reaction $\text{N}_2\text{(g)}+3\text{H}_2\text{(g)}\rightleftharpoons2\text{NH}_3\text{(g)}$.

Show answer
There are $4$ moles of gas on the reactant side but only $2$ moles of gas on the product side. Since gases have much greater entropy (disorder) than the same substances would in a more ordered state, and fewer gas particles means fewer possible microstates (less disorder), the entropy of the system decreases, giving a negative $\Delta S^{\ominus}$.
[2]

Section C – Electrochemistry

SL and HL
9

A voltaic cell is set up using the half-cells $\text{Zn}^{2+}\text{(aq)}+2e^-\rightleftharpoons\text{Zn(s)}$, $E^{\ominus}=-0.76\,\text{V}$, and $\text{Cu}^{2+}\text{(aq)}+2e^-\rightleftharpoons\text{Cu(s)}$, $E^{\ominus}=+0.34\,\text{V}$.(a) Calculate the standard cell potential, $E_{\text{cell}}^{\ominus}$.(b) State which electrode is the anode (oxidation) and which is the cathode (reduction).

Show answer
(a) $E_{\text{cell}}^{\ominus}=E^{\ominus}_{\text{cathode}}-E^{\ominus}_{\text{anode}}=0.34-(-0.76)=+1.10\,\text{V}$.(b) Zinc has the more negative electrode potential, so it is oxidised at the anode ($\text{Zn}\rightarrow\text{Zn}^{2+}+2e^-$); copper is reduced at the cathode ($\text{Cu}^{2+}+2e^-\rightarrow\text{Cu}$).
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10

Using $\Delta G^{\ominus}=-nFE_{\text{cell}}^{\ominus}$, calculate the standard Gibbs energy change, in $\text{kJ mol}^{-1}$, for the cell reaction in the previous question ($n=2$).

Show answer
$\Delta G^{\ominus}=-nFE_{\text{cell}}^{\ominus}=-2(96500)(1.10)=-212300\,\text{J mol}^{-1}=-212\,\text{kJ mol}^{-1}$.
[3]
11

Explain the relationship between the sign of $E_{\text{cell}}^{\ominus}$ and the sign of $\Delta G^{\ominus}$, and state what this tells you about whether a redox reaction is spontaneous.

Show answer
Since $\Delta G^{\ominus}=-nFE_{\text{cell}}^{\ominus}$ and $n$ and $F$ are always positive, a positive $E_{\text{cell}}^{\ominus}$ always corresponds to a negative $\Delta G^{\ominus}$, and vice versa. A negative $\Delta G^{\ominus}$ (positive $E_{\text{cell}}^{\ominus}$) means the reaction is thermodynamically spontaneous (feasible) under standard conditions.
[2]
12

The enthalpy of solution of an ionic compound can be found from a Hess’s law cycle linking lattice enthalpy and enthalpy of hydration. Describe this cycle, and explain, in terms of these two enthalpy changes, why some ionic compounds dissolve exothermically while others dissolve endothermically.

Show answer
In the cycle, the ionic solid is first (hypothetically) converted into its gaseous ions by supplying the lattice enthalpy (the reverse of lattice formation, which is endothermic), and these gaseous ions are then hydrated by surrounding water molecules, releasing the (exothermic) enthalpies of hydration; by Hess’s law, $\Delta H_{\text{solution}}=-\Delta H_{\text{lattice}}+\Delta H_{\text{hydration(cation)}}+\Delta H_{\text{hydration(anion)}}$. If the total (negative) hydration enthalpy is larger in magnitude than the lattice enthalpy, dissolution is exothermic overall; if the lattice enthalpy is larger in magnitude than the total hydration enthalpy, dissolution is endothermic overall.
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