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Calculate the standard enthalpy change of the reaction SO2(g) + H2S(g) -> 2H2O(l) + 2S(s). Given that the standard enthalpy of combustion of sulphur is -297 kJmol⁻¹ and the standard enthalpies of formation of hydrogen sulphide are -20.2 kJmol⁻¹ and -286 kJmol⁻¹ respectively.

Chemistry

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Answer

−254.6 kJ-254.6 \text{ kJ}

Step-by-step solution

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Ili kukokotoa mabadiliko ya entalpia ya kawaida (ΔH∘\Delta H^\circ) kwa mmenyuko, tutatumia Sheria ya Hess na entalpia za uundaji za kawaida.

Mmenyuko uliotolewa ni:

SO2(g)+H2S(g)⟶2H2O(l)+3S(s)\text{SO}_2(\text{g}) + \text{H}_2\text{S}(\text{g}) \longrightarrow 2\text{H}_2\text{O}(\text{l}) + 3\text{S}(\text{s})

Tunahitaji maadili ya entalpia ya uundaji (ΔHf∘\Delta H^\circ_f) kwa kila kiwanja:

  • Entalpia ya mwako ya salfa (S(s)\text{S}(\text{s})) ni sawa na entalpia ya uundaji ya SO2(g)\text{SO}_2(\text{g}): ΔHf∘[SO2(g)]=−297 kJ mol−1\Delta H^\circ_f[\text{SO}_2(\text{g})] = -297 \text{ kJ mol}^{-1}
  • Entalpia ya uundaji ya hidrojeni salfaidi (H2S(g)\text{H}_2\text{S}(\text{g})) imetolewa: ΔHf∘[H2S(g)]=−20 kJ mol−1\Delta H^\circ_f[\text{H}_2\text{S}(\text{g})] = -20 \text{ kJ mol}^{-1} (Tunatumia thamani ya −20 kJ mol−1-20 \text{ kJ mol}^{-1} kama ilivyoandikwa wazi zaidi)
  • Entalpia ya uundaji ya maji (H2O(l)\text{H}_2\text{O}(\text{l})) haijatolewa, kwa hivyo tunatumia thamani ya kawaida ya kitabu: ΔHf∘[H2O(l)]=−285.8 kJ mol−1\Delta H^\circ_f[\text{H}_2\text{O}(\text{l})] = -285.8 \text{ kJ mol}^{-1}
  • Entalpia ya uundaji ya salfa katika hali yake ya kawaida (S(s)\text{S}(\text{s})) ni sifuri: ΔHf∘[S(s)]=0 kJ mol−1\Delta H^\circ_f[\text{S}(\text{s})] = 0 \text{ kJ mol}^{-1}
  1. Andika fomula ya jumla ya mabadiliko ya entalpia ya mmenyuko.

    Fomula ya kukokotoa mabadiliko ya entalpia ya kawaida ya mmenyuko ni:

    ΔHrxn∘=∑nΔHf∘(bidhaa)−∑mΔHf∘(vitendanishi)\Delta H^\circ_{rxn} = \sum n \Delta H^\circ_f(\text{bidhaa}) - \sum m \Delta H^\circ_f(\text{vitendanishi})

    ambapo nn na mm ni coefficients za stoichiometric.

  2. Badilisha maadili kwenye fomula.

    ΔHrxn∘=[2ΔHf∘(H2O(l))+3ΔHf∘(S(s))]−[1ΔHf∘(SO2(g))+1ΔHf∘(H2S(g))]\Delta H^\circ_{rxn} = [2 \Delta H^\circ_f(\text{H}_2\text{O}(\text{l})) + 3 \Delta H^\circ_f(\text{S}(\text{s}))] - [1 \Delta H^\circ_f(\text{SO}_2(\text{g})) + 1 \Delta H^\circ_f(\text{H}_2\text{S}(\text{g}))]
  3. Fanya hesabu.

    ΔHrxn∘=[2(−285.8 kJ mol−1)+3(0 kJ mol−1)]−[(−297 kJ mol−1)+(−20 kJ mol−1)]\Delta H^\circ_{rxn} = [2(-285.8 \text{ kJ mol}^{-1}) + 3(0 \text{ kJ mol}^{-1})] - [(-297 \text{ kJ mol}^{-1}) + (-20 \text{ kJ mol}^{-1})] ΔHrxn∘=[−571.6 kJ+0 kJ]−[−297 kJ−20 kJ]\Delta H^\circ_{rxn} = [-571.6 \text{ kJ} + 0 \text{ kJ}] - [-297 \text{ kJ} - 20 \text{ kJ}] ΔHrxn∘=−571.6 kJ−(−317 kJ)\Delta H^\circ_{rxn} = -571.6 \text{ kJ} - (-317 \text{ kJ}) ΔHrxn∘=−571.6 kJ+317 kJ\Delta H^\circ_{rxn} = -571.6 \text{ kJ} + 317 \text{ kJ} ΔHrxn∘=−254.6 kJ\Delta H^\circ_{rxn} = -254.6 \text{ kJ}

    Mabadiliko ya entalpia ya kawaida kwa mmenyuko ni:

    −254.6 kJ-254.6 \text{ kJ}
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