Kalorimetri, rute Hess, dan batas energi ikatan
Diperkenalkan| English | Makna |
|---|---|
| enthalpy/enˈθælpi/ | entalpi |
| exothermic/eɡzəˈðɜːmɪk/ | eksoterm |
| endothermic/ˌendəʊˈθɜːmɪk/ | endotermik |
| Hess's law/ˈhesɪz lɔː/ | Hukum Hess |
| mean bond enthalpy | entalpi ikatan rata-rata |
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.
Entalpi pembentukan standar selalu membentuk satu mol produk dalam keadaan gas.
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Pembentukan menghasilkan satu mol senyawa dalam keadaan standar yang ditentukan. Pembentukan air cair menghasilkan air cair; atom-atom gas termasuk dalam atomisasi.
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 larutan naik sebesar 4.00°C. Gunakan c = 4.18 J per gram per °C. Hitung energi yang diserap oleh larutan dalam Joule.
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q = mcΔT = 100.0 × 4.18 × 4.00 = 1672 J. Peningkatan panas positif ini dimiliki oleh larutan; panas reaksi memiliki tanda berlawanan.
80.0 g larutan meningkat sebesar 4.50°C ketika 0.0300 mol bereaksi secara eksotermik. Gunakan c = 4.18 J per gram per °C. Hitung entalpi reaksi dalam 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. Bagi dengan mol yang bereaksi dan konversi J ke 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 dan 2H₂ terbakar dengan entalpi gabungan −966 kJ per mol CH₄ yang terbentuk; pembakaran CH₄ adalah −891 kJ per mol. Nilai pembentukan mana yang benar?
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Kedua rute mencapai produk yang sama: ΔH + (−891) = −966, sehingga ΔH = −75 kJ per mol. Menambahkan dua bilangan negatif mengikuti arah rute yang tidak konsisten.
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.
Untuk H₂(g) + Cl₂(g) → 2HCl(g), ΔH = −184 kJ per mol. Ikatan H–H = 436 dan Cl–Cl = 242 kJ per mol. Perkirakan entalpi ikatan H–Cl dalam kJ per mol.
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ΔH = diputus − terbentuk: −184 = 436 + 242 − 2x. Maka 2x = 862 dan x = 431 kJ per mol. Terdapat dua ikatan H–Cl pada produk.