Calorimetry, Hess routes and bond-energy limits
Introduced| English | Meaning |
|---|---|
| enthalpy/enˈθælpi/ | A state function equal to internal energy plus the product of pressure and volume. |
| exothermic/eɡzəˈðɜːmɪk/ | Describing a process that transfers heat to its surroundings. |
| endothermic/ˌendəʊˈθɜːmɪk/ | Describing a process that takes in heat from its surroundings. |
| Hess's law/ˈhesɪz lɔː/ | The enthalpy change between given initial and final states is the same whatever sequence of steps connects them. |
| mean bond enthalpy | The average enthalpy change for breaking one mole of a specified covalent bond type homolytically in gaseous molecules. |
Why does the same reaction give different temperature rises?
- A reaction can warm a small cup strongly and a larger cup only slightly. Temperature rise alone does not give the energy per mole: track the heated mass and the amount reacting.
- At constant pressure, enthalpy 焓 change is the heat change of the reaction. Standard conditions mean 100 kPa and a specified temperature, usually 298 K, with substances in their standard states.
Define the reaction before choosing its sign
- Exothermic 放热的 reactions release energy to the surroundings and have negative ΔH; endothermic 吸热的 reactions take energy from the surroundings and have positive ΔH. Draw a vertical enthalpy axis, labelled reactant/product levels and a signed arrow between them.
- Standard reaction enthalpy refers to the equation as written. Formation makes one mole of a compound from its elements in standard states; combustion burns one mole completely in oxygen; neutralisation forms one mole of water from acid and alkali; atomisation forms one mole of gaseous atoms from the element in its standard state.
Standard formation enthalpy always forms one mole of gaseous product.
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Formation makes one mole of the compound in its specified standard state. Liquid water formation produces liquid water; gaseous atoms belong to atomisation.
Convert a corrected temperature rise into energy per mole
- For a solution, use $q=mc\Delta T$. If 80.0 g has $c=4.18\,\dfrac{\mathrm{J}}{\mathrm{g}\,{}^\circ\mathrm C}$ and a corrected rise of 5.50°C, $q=80.0\times4.18\times5.50=1839.2\,\mathrm J$. This is energy gained by the solution.
- If 0.0400 mol reacts, $\Delta H=-\dfrac{1839.2}{1000\times0.0400}=-46.0\,\dfrac{\mathrm{kJ}}{\mathrm{mol}}$. Use the reacting amount required by the equation, convert J to kJ, and give the reaction the opposite sign to the solution's heat gain.
A mixture starts at 20.0°C. Mixing occurs at 120 s; readings are 25.2°C at 150 s and 24.9°C at 180 s. The straight cooling gradient is −0.010°C/s. Extrapolation to 120 s gives 25.5°C, so the corrected rise is 5.5°C. This model corrects estimated cooling after mixing; it does not automatically correct evaporation or every heat-capacity assumption.
100.0 g of solution rises by 4.00°C. Use c = 4.18 J per gram per °C. Calculate energy gained by the solution in J.
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q = mcΔT = 100.0 × 4.18 × 4.00 = 1672 J. This positive heat gain belongs to the solution; reaction heat has the opposite sign.
80.0 g of solution rises by 4.50°C when 0.0300 mol reacts exothermically. Use c = 4.18 J per gram per °C. Calculate reaction enthalpy in kJ per mol.
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q = 80.0 × 4.18 × 4.50 = 1504.8 J; ΔH = −1504.8/(1000 × 0.0300) = −50.16 kJ per mol. Divide by reacting moles and convert J to kJ.
Join routes with Hess's law
- Hess's law 赫斯定律 says total enthalpy change depends on the initial and final states, not the route. Reversing an equation reverses ΔH; multiplying all coefficients multiplies ΔH. Cancel identical species in the same states.
- For C(s) + 2H₂(g) → CH₄(g), supplied combustion values are −394, −286 and −891 $\dfrac{\mathrm{kJ}}{\mathrm{mol}}$ for C, H₂ and CH₄. Both routes end at CO₂(g) + 2H₂O(l): $\Delta H+(-891)=-394+2(-286)$, giving $\Delta H=-75\,\dfrac{\mathrm{kJ}}{\mathrm{mol}}$.
C and 2H₂ combust with combined enthalpy −966 kJ per mol of CH₄ formed; CH₄ combustion is −891 kJ per mol. Which formation value follows?
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Both routes reach the same products: ΔH + (−891) = −966, hence ΔH = −75 kJ per mol. Adding the two negative numbers follows inconsistent route directions.
Use bond data within its limits
- Breaking bonds takes energy; making bonds releases it. For gaseous H₂ + Cl₂ → 2HCl with supplied bond enthalpies 436, 242 and 431 $\dfrac{\mathrm{kJ}}{\mathrm{mol}}$, estimate $\Delta H=(436+242)-2(431)=-184\,\dfrac{\mathrm{kJ}}{\mathrm{mol}}$.
- A mean bond enthalpy 平均键焓 averages that bond in different gaseous compounds. Rearranging the same equation can estimate an unknown bond value. Liquid-water formation needs phase-change data; a weaker bond can suggest easier breaking but does not determine an exact rate by itself.
Heat loss usually makes a measured exothermic rise too small and ΔH less negative. In combustion, incomplete burning and evaporation can introduce other errors. State the actual experiment, assumptions and direction of the effect. Written calorimetry and Hess reasoning do not certify supervised practical performance.
Write the equation and states, identify the heated material, correct the temperature where justified, calculate q and divide by the appropriate mole amount. Check sign, units and the limits of the data.
For gaseous H₂ + Cl₂ → 2HCl, ΔH = −184 kJ per mol. H–H = 436 and Cl–Cl = 242 kJ per mol. Estimate H–Cl bond enthalpy in kJ per mol.
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ΔH = broken − formed: −184 = 436 + 242 − 2x. Thus 2x = 862 and x = 431 kJ per mol. There are two H–Cl bonds in the products.