DSE Chemistry Diagnostic: Energetics / Thermochemistry

Unit Test 1: Hess's Law Cycle Construction

Question

Given the following standard enthalpy changes:

Reaction	$\Delta H$ (kJ/mol)
$C(s) + O_{2}(g) \rightarrow CO_{2}(g)$	$-393.5$
$H_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow H_{2}O(l)$	$-285.8$
$CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l)$	$-890.3$

(a) Using Hess's law, calculate the standard enthalpy of formation of methane, $\Delta H_{f}^{\circ}(CH_{4})$. [4 marks]

(b) Draw a Hess's law cycle that clearly shows the two alternative routes from elements to products. [2 marks]

(c) A student writes $\Delta H_{f}^{\circ}(CH_{4}) = -393.5 - 285.8 - (-890.3) = +210.6$ kJ/mol. Identify the error in the student's working. [2 marks]

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Worked Solution

(a) Target: $C(s) + 2H_{2}(g) \rightarrow CH_{4}(g)$

Using the cycle: elements $\rightarrow$ products (via elements $\rightarrow$ methane $\rightarrow$ products or directly).

$\Delta H_{f}^{\circ}(CH_{4}) + \Delta H_{c}^{\circ}(CH_{4}) = \Delta H_{f}^{\circ}(CO_{2}) + 2\Delta H_{f}^{\circ}(H_{2}O)$

$\Delta H_{f}^{\circ}(CH_{4}) + (-890.3) = (-393.5) + 2 \times (-285.8)$

$\Delta H_{f}^{\circ}(CH_{4}) - 890.3 = -393.5 - 571.6$

$\Delta H_{f}^{\circ}(CH_{4}) - 890.3 = -965.1$

$\Delta H_{f}^{\circ}(CH_{4}) = -965.1 + 890.3 = -74.8 \text{ kJ/mol}$

(b) Hess's cycle:

C(s) + 2H2(g) --[ΔHf°(CH4)]--> CH4(g)
     |                                |
     |                                | [ΔHc°(CH4)]
     |                                v
     +---[ΔHf°(CO2) + 2ΔHf°(H2O)]--> CO2(g) + 2H2O(l)

Route 1: Elements $\rightarrow$ Methane $\rightarrow$ Products: $\Delta H_{f}^{\circ}(CH_{4}) + \Delta H_{c}^{\circ}(CH_{4})$

Route 2: Elements $\rightarrow$ Products (directly): $\Delta H_{f}^{\circ}(CO_{2}) + 2\Delta H_{f}^{\circ}(H_{2}O)$

(c) The student's arithmetic is correct ($-393.5 - 571.6 + 890.3 = -74.8$), but the numerical answer they obtained ($+210.6$) is wrong because they only subtracted one $\Delta H_{f}^{\circ}(H_{2}O)$ instead of two. The combustion of methane produces two moles of water, so $2 \times (-285.8)$ must be used. The student calculated: $-393.5 - 285.8 + 890.3 = +211.0$ (not even their stated answer of 210.6). The correct calculation requires the factor of 2.

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Unit Test 2: Average Bond Enthalpy vs Exact

Question

The standard enthalpy change for the reaction $H_{2}(g) + Cl_{2}(g) \rightarrow 2HCl(g)$ is $-184.6$ kJ/mol.

(a) Using average bond enthalpies ($H-H = 436$ kJ/mol, $Cl-Cl = 242$ kJ/mol, $H-Cl = 431$ kJ/mol), calculate the enthalpy change for this reaction. [3 marks]

(b) Explain why the value obtained in (a) differs from the actual value of $-184.6$ kJ/mol. [3 marks]

(c) Which type of calculation is more accurate: using average bond enthalpies or using standard enthalpy of formation data? Explain. [2 marks]

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Worked Solution

(a) Bonds broken:

$H-H: 436 \text{ kJ/mol}$ $Cl-Cl: 242 \text{ kJ/mol}$ $\text{Total} = 436 + 242 = 678 \text{ kJ/mol}$

Bonds formed:

$2 \times H-Cl = 2 \times 431 = 862 \text{ kJ/mol}$

$\Delta H = 678 - 862 = -184 \text{ kJ/mol}$

(b) The calculated value ($-184$) is very close to the actual value ($-184.6$), but the small difference arises because:

1. Average bond enthalpies are averages taken over many different compounds containing that bond type. The actual $H-Cl$ bond enthalpy in $HCl(g)$ may differ slightly from the average value.
2. Average bond enthalpies are typically given for species in the gaseous state at 298 K and refer to the mean bond dissociation enthalpy, which may differ from the specific bond enthalpy in this particular reaction.
3. The bond enthalpy values have experimental uncertainties.

(c) Using standard enthalpy of formation data is more accurate because these are experimentally measured values specific to each compound, rather than averages. Average bond enthalpies introduce error because the actual bond energy in a specific molecule can deviate from the average due to the molecular environment.

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Unit Test 3: Calorimetry with Heat Loss

Question

50.0 cm$^{3}$ of 1.00 mol/dm$^{3}$ $HCl$ was mixed with 50.0 cm$^{3}$ of 1.00 mol/dm$^{3}$ $NaOH$ in a polystyrene cup. The temperature rose from 20.0$^{\circ}$C to 26.7$^{\circ}$C.

(a) Calculate the enthalpy change of neutralisation per mole of water formed. [4 marks]

(b) The theoretical value is $-57.1$ kJ/mol. Calculate the percentage error and suggest a reason for any discrepancy. [2 marks]

(c) A student claims that using a glass beaker instead of a polystyrene cup would give a more accurate result. Evaluate this claim. [2 marks]

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Worked Solution

(a) Assume density of solution = 1.00 g/cm$^{3}$ and specific heat capacity = 4.18 J/(g$\cdot$K).

Total volume = $50.0 + 50.0 = 100.0$ cm$^{3}$

Mass of solution = $100.0$ g

Temperature change = $26.7 - 20.0 = 6.7$ K

$q = mc\Delta T = 100.0 \times 4.18 \times 6.7 = 2800.6 \text{ J} = 2.80 \text{ kJ}$

Moles of water formed:

$n(HCl) = 1.00 \times 0.0500 = 0.0500 \text{ mol}$ $n(NaOH) = 1.00 \times 0.0500 = 0.0500 \text{ mol}$ $n(H_{2}O) = 0.0500 \text{ mol} \text{ (limiting reagent = both, equimolar)}$

$\Delta H = \frac{-q}{n} = \frac{-2.80}{0.0500} = -56.0 \text{ kJ/mol}$

(b) $\%\text{ error} = \frac{|-56.0 - (-57.1)|}{57.1} \times 100\% = \frac{1.1}{57.1} \times 100\% = 1.9\%$

The discrepancy is due to heat loss to the surroundings (the polystyrene cup is not a perfect insulator) and the heat absorbed by the cup itself (which was not accounted for in the calculation).

(c) The claim is incorrect. A glass beaker is a much better conductor of heat than a polystyrene cup, so more heat would be lost to the surroundings. This would result in a smaller temperature rise and a less exothermic (less negative) $\Delta H$ value, increasing the error.

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Integration Test 1: Hess's Law + Bond Enthalpy Comparison

Question

Given:

• $\Delta H_{f}^{\circ}(C_{2}H_{4}, g) = +52.2$ kJ/mol
• $\Delta H_{f}^{\circ}(C_{2}H_{6}, g) = -84.7$ kJ/mol
• Bond enthalpies: $C=C = 612$, $C-C = 348$, $C-H = 412$, $H-H = 436$ (all in kJ/mol)

(a) Calculate the enthalpy change for the hydrogenation of ethene using standard enthalpy of formation data:

$C_{2}H_{4}(g) + H_{2}(g) \rightarrow C_{2}H_{6}(g)$

[2 marks]

(b) Calculate the same enthalpy change using average bond enthalpies. [4 marks]

(c) Account for any difference between the two values obtained in (a) and (b). [3 marks]

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Worked Solution

(a) $\Delta H_{\text{rxn}} = \Delta H_{f}^{\circ}(C_{2}H_{6}) - \Delta H_{f}^{\circ}(C_{2}H_{4}) - \Delta H_{f}^{\circ}(H_{2})$

$= -84.7 - 52.2 - 0 = -136.9 \text{ kJ/mol}$

(b) Bonds broken:

• $C=C$: 612 kJ/mol
• $H-H$: 436 kJ/mol In $C_{2}H_{4}$ ($H_{2}C=CH_{2}$): there are $1 \times C=C$ and $4 \times C-H$ bonds. In $H_{2}$: $1 \times H-H$.

Bonds broken: $C=C + H-H + 4 \times C-H = 612 + 436 + 1648 = 2696$ kJ/mol

Bonds formed (in $C_{2}H_{6}$, $H_{3}C-CH_{3}$): $1 \times C-C + 6 \times C-H = 348 + 6 \times 412 = 348 + 2472 = 2820$ kJ/mol

$\Delta H = 2696 - 2820 = -124 \text{ kJ/mol}$

(c) The value from formation data ($-136.9$ kJ/mol) differs from the bond enthalpy value ($-124$ kJ/mol) by 12.9 kJ/mol.

Reasons:

1. Average bond enthalpies are averaged over many different compounds. The actual $C-H$ bond enthalpy in ethane differs from the average $C-H$ bond enthalpy because the electronic environment around each $C-H$ bond varies.
2. The $C=C$ bond enthalpy in ethene specifically may differ from the average $C=C$ bond enthalpy.
3. The bond enthalpy of $H-H$ in $H_{2}$ gas is well-defined, but the $C-C$ and $C-H$ values are averages.

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Integration Test 2: Born-Haber-Type Cycle

Question

The enthalpy of formation of magnesium oxide is $-601.6$ kJ/mol.

Given:

• Enthalpy of atomisation of Mg: $+147.1$ kJ/mol
• First ionisation energy of Mg: $+738$ kJ/mol
• Second ionisation energy of Mg: $+1451$ kJ/mol
• Bond dissociation enthalpy of $O_{2}$: $+498$ kJ/mol
• First electron affinity of O: $-141$ kJ/mol
• Second electron affinity of O: $+798$ kJ/mol

(a) Construct a Born-Haber cycle for the formation of $MgO(s)$ from its elements. [3 marks]

(b) Use the Born-Haber cycle to calculate the lattice enthalpy of $MgO(s)$. [3 marks]

(c) Explain why the second electron affinity of oxygen is endothermic. [2 marks]

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Worked Solution

(a) Born-Haber cycle:

Mg(s) + 1/2 O2(g) --[ΔHf°]--> MgO(s)
     |                              |
     | +147.1                       | [ΔHlattice]
     v                              v
Mg(g) + 1/2 O2(g)                  Mg2+(g) + O2-(g)
     |                              ^
     | +738, +1451                  |
     v                              | -141, +798
Mg2+(g) + 1/2 O2(g)                 |
     |                              |
     | +249 (1/2 x 498)             |
     v                              |
Mg2+(g) + O(g) ---------------------+

(b) Hess's law:

$\Delta H_{f}^{\circ} = \Delta H_{\text{atom}}(Mg) + IE_{1} + IE_{2} + \frac{1}{2}\Delta H_{\text{diss}}(O_{2}) + EA_{1} + EA_{2} + \Delta H_{\text{lattice}}$

$-601.6 = 147.1 + 738 + 1451 + 249 + (-141) + 798 + \Delta H_{\text{lattice}}$

$-601.6 = 147.1 + 738 + 1451 + 249 - 141 + 798 + \Delta H_{\text{lattice}}$

Sum of all terms except lattice: $147.1 + 738 + 1451 + 249 - 141 + 798 = 3242.1$

$-601.6 = 3242.1 + \Delta H_{\text{lattice}}$

$\Delta H_{\text{lattice}} = -601.6 - 3242.1 = -3843.7 \text{ kJ/mol}$

(c) The second electron affinity of oxygen is endothermic because the $O^{-}$ ion already carries a negative charge. Adding a second electron to $O^{-}$ requires overcoming electrostatic repulsion between the incoming electron and the existing negative charge. Energy must be supplied to force the second electron onto the negatively charged ion.

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Integration Test 3: Multi-Step Energetics

Question

Methanol ($CH_{3}OH$) can be used as a fuel. The complete combustion of methanol is:

$2CH_{3}OH(l) + 3O_{2}(g) \rightarrow 2CO_{2}(g) + 4H_{2}O(l) \quad \Delta H = -1452 \text{ kJ/mol (for the equation as written)}$

The standard enthalpy of formation of $CO_{2}(g)$ is $-393.5$ kJ/mol and of $H_{2}O(l)$ is $-285.8$ kJ/mol.

(a) Calculate the standard enthalpy of formation of methanol, $CH_{3}OH(l)$. [3 marks]

(b) Calculate the enthalpy change when 10.0 g of methanol is burned completely. [2 marks]

(c) When 10.0 g of methanol was burned to heat 500 g of water, the temperature of the water rose by 31.2$^{\circ}$C. Calculate the experimental enthalpy of combustion per mole of methanol and the percentage efficiency of the experiment. [4 marks]

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Worked Solution

(a) For 2 mol of methanol:

$\Delta H_{\text{rxn}} = 2\Delta H_{f}^{\circ}(CO_{2}) + 4\Delta H_{f}^{\circ}(H_{2}O) - 2\Delta H_{f}^{\circ}(CH_{3}OH) - 3\Delta H_{f}^{\circ}(O_{2})$

$-1452 = 2(-393.5) + 4(-285.8) - 2\Delta H_{f}^{\circ}(CH_{3}OH) - 0$

$-1452 = -787.0 - 1143.2 - 2\Delta H_{f}^{\circ}(CH_{3}OH)$

$-1452 = -1930.2 - 2\Delta H_{f}^{\circ}(CH_{3}OH)$

$2\Delta H_{f}^{\circ}(CH_{3}OH) = -1930.2 + 1452 = -478.2$

$\Delta H_{f}^{\circ}(CH_{3}OH) = -239.1 \text{ kJ/mol}$

(b) $M(CH_{3}OH) = 12.0 + 4 \times 1.0 + 16.0 + 1.0 = 32.0 \text{ g/mol}$

$n(CH_{3}OH) = \frac{10.0}{32.0} = 0.3125 \text{ mol}$

Enthalpy change per mole = $\frac{-1452}{2} = -726$ kJ/mol

$\Delta H = 0.3125 \times (-726) = -226.9 \text{ kJ}$

(c) Heat absorbed by water:

$q = mc\Delta T = 500 \times 4.18 \times 31.2 = 65208 \text{ J} = 65.2 \text{ kJ}$

Experimental enthalpy of combustion per mole:

$\Delta H_{\text{exp}} = \frac{-65.2}{0.3125} = -208.6 \text{ kJ/mol}$

Percentage efficiency:

$\%\text{ efficiency} = \frac{208.6}{726} \times 100\% = 28.7\%$

This low efficiency is due to significant heat loss to the surroundings, incomplete combustion, and heat absorbed by the container.