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12 November, 23:11

Consider the reaction:

2H2O (l) 2H2 (g) + O2 (g)

Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.73 moles of H2O (l) react at standard conditions.

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  1. 12 November, 23:13
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    The correct answer is - 1659.17 J/K.

    Explanation:

    The reaction given is:

    2H₂O (l) ⇔ 2H₂ (g) + O₂ (g)

    In the given case, first there is a need to find ΔHreaction, which is equivalent to ΔHf (products) - ΔHf (reactants)

    Based on the standard thermodynamic table, the ΔHf (H₂O) is - 285.8 KJ/mol, the ΔHf (H₂) is 0 KJ/mol, and the ΔHf (O₂) is 0 KJ/mol.

    On putting the values, the ΔHreaction will be,

    ΔHreaction = 2 * ΔHf (H₂) + ΔHf (O₂) - 2 * ΔHf (H₂O)

    = 2 * 0 + 0 - 2 * (-285.8 KJ/mol) = 571.6 KJ

    The calculated value of ΔHreaction is for the two moles of H₂O, now for 1.73 moles of H₂O it will be,

    ΔHreaction = + 571.6 KJ / 2 mol * 1.73 mol = 494.434 KJ

    The temperature given in the question is 298 K, now ΔSsurrounding will be,

    ΔSsurrounding = - ΔHreaction/T = - 494434 J/298 K = - 1659.17 J/K.
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