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The standard entropy change for the reaction
4Fe(s) + 3O2(g) → 2Fe2O3(s) is -550 JK-1 at 298 K.
[Given: The standard enthalpy change for the reaction is -165 kJ mol-1]. The temperature in K at which the reaction attains equilibrium is __________. (Nearest Integer)
    Correct answer is '300'. Can you explain this answer?
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    The standard entropy change for the reaction4Fe(s) + 3O2(g)→2Fe2O...

    = 300 K
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    The standard entropy change for the reaction4Fe(s) + 3O2(g)→2Fe2O...
    The standard entropy change for the reaction 4Fe(s) + 3O2(g) can be calculated by subtracting the sum of the standard entropy values of the reactants from the sum of the standard entropy values of the products.

    The standard entropy values for Fe(s) and O2(g) can be found in tables or literature. Let's assume the standard entropy values are as follows:

    Fe(s) = 27.3 J/(mol·K)
    O2(g) = 205.0 J/(mol·K)

    The standard entropy change for the reaction can be calculated as:

    ΔS° = (4 mol Fe × 27.3 J/(mol·K)) + (3 mol O2 × 205.0 J/(mol·K)) - (0 mol Fe + 0 mol O2)

    ΔS° = (4 × 27.3) + (3 × 205.0) - (0 + 0)
    ΔS° = 109.2 + 615.0 - 0
    ΔS° = 724.2 J/K

    Therefore, the standard entropy change for the reaction 4Fe(s) + 3O2(g) is 724.2 J/K.
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    The standard entropy change for the reaction4Fe(s) + 3O2(g)→2Fe2O3(s) is -550 JK-1at 298 K.[Given: The standard enthalpy change for the reaction is -165 kJ mol-1]. The temperature in K at which the reaction attains equilibrium is __________. (Nearest Integer)Correct answer is '300'. Can you explain this answer?
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