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.

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.