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16 April, 05:57

For the process 2SO2 (g) + O2 (g) Right arrow. 2SO3 (g), Delta. S = - 187.9 J/K and Delta. H = - 198.4 kJ at 297.0 K are known. What is the entropy of this reaction? Use Delta. G = Delta. H - TDelta. S.

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  1. 16 April, 07:49
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    The entropy of this reaction (ΔS) = - 187.9 J/K

    Explanation:

    . Step : Data given

    ΔS = - 187.9 J/K

    ΔH = - 198.4 kJ

    T = 297.0 K

    Step 2: The balanced equation

    2SO2 (g) + O2 (g) → 2SO3 (g)

    Step 3: Calculate the entropy

    ΔG = ΔH - TΔS

    ⇒ΔG = the gibbs free energy

    ⇒ΔH = the change in enthalpy of the reaction = - 198.4 kJ

    ⇒with T = the temperature = 297.0 K

    ⇒with ΔS = the change of entropy = - 187.9 J/K

    ΔG = - 198400 kJ - 297 * (-187.9kJ/mol)

    ΔG = - 198400+55806.3

    ΔG = - 142593.7 kJ = the gibbs free energy

    The entropy of this reaction (ΔS) = - 187.9 J/K
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