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.  

at the bottom

explanation:

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.