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Calculate the enthalpy change for the following reaction: C2H4(g) + Br2(g) -> C2H4Br2(g)

Chemistry

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Answer

-103 kJ/mol\text{-103 kJ/mol}

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Ili kukokotoa mabadiliko ya enthalpy (ΔH\Delta H) kwa mmenyuko, tunatumia fomula ifuatayo:

ΔH=∑(nishati ya vifungo vilivyovunjwa)−∑(nishati ya vifungo vilivyoumbwa)\Delta H = \sum (\text{nishati ya vifungo vilivyovunjwa}) - \sum (\text{nishati ya vifungo vilivyoumbwa})

Mmenyuko ni: C2H4(g)+Br2(g)⟶C2H4Br2(g)\text{C}_2\text{H}_4(\text{g}) + \text{Br}_2(\text{g}) \longrightarrow \text{C}_2\text{H}_4\text{Br}_2(\text{g})

  1. Tambua vifungo vilivyovunjwa katika vitendanishi.

    Katika C2H4\text{C}_2\text{H}_4 (ethene), kuna kifungo kimoja cha C=C na vifungo vinne vya C-H.
    Katika Br2\text{Br}_2, kuna kifungo kimoja cha Br-Br.
    Vifungo vinne vya C-H hubaki bila kubadilika katika bidhaa, kwa hivyo havijumuishwi katika hesabu ya mabadiliko ya enthalpy.
    Vifungo vilivyovunjwa ni:

    • Kifungo 1 cha C=C
    • Kifungo 1 cha Br-Br

  2. Tambua vifungo vilivyoumbwa katika bidhaa.

    Katika C2H4Br2\text{C}_2\text{H}_4\text{Br}_2 (1,2-dibromoethane), kifungo cha C=C hubadilika na kuwa kifungo kimoja cha C-C, na vifungo viwili vipya vya C-Br huundwa.
    Vifungo vilivyoumbwa ni:

    • Kifungo 1 cha C-C
    • Vifungo 2 vya C-Br

  3. Hesabu jumla ya nishati inayohitajika kuvunja vifungo.

    Kutoka kwenye jedwali:
    Nishati ya C=C = 610 kJ/mol610 \text{ kJ/mol}
    Nishati ya Br-Br = 193 kJ/mol193 \text{ kJ/mol}
    Jumla ya nishati inayohitajika kuvunja vifungo = (1×610 kJ/mol)+(1×193 kJ/mol)(1 \times 610 \text{ kJ/mol}) + (1 \times 193 \text{ kJ/mol})

    =610 kJ/mol+193 kJ/mol=803 kJ/mol= 610 \text{ kJ/mol} + 193 \text{ kJ/mol} = 803 \text{ kJ/mol}

  4. Hesabu jumla ya nishati inayotolewa wakati vifungo vipya vinaundwa.

    Kutoka kwenye jedwali:
    Nishati ya C-C = 346 kJ/mol346 \text{ kJ/mol}
    Nishati ya C-Br = 280 kJ/mol280 \text{ kJ/mol}
    Jumla ya nishati inayotolewa wakati vifungo vipya vinaundwa = (1×346 kJ/mol)+(2×280 kJ/mol)(1 \times 346 \text{ kJ/mol}) + (2 \times 280 \text{ kJ/mol})

    =346 kJ/mol+560 kJ/mol=906 kJ/mol= 346 \text{ kJ/mol} + 560 \text{ kJ/mol} = 906 \text{ kJ/mol}

  5. Hesabu mabadiliko ya enthalpy (ΔH\Delta H).

    ΔH=803 kJ/mol−906 kJ/mol\Delta H = 803 \text{ kJ/mol} - 906 \text{ kJ/mol} ΔH=−103 kJ/mol\Delta H = -103 \text{ kJ/mol}

    Mabadiliko ya enthalpy kwa mmenyuko ni:

    -103 kJ/mol\text{-103 kJ/mol}
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