Calculate the entropy change when 3.66 g of h2 reacts with o2 according to the reaction 2 h2(g) + o2(g) → 2 h2o(ℓ) at 298 k and 1 atm pressure. the standard molar enthalpy of formation of h2o(ℓ) at 298 k is −285.8 kj/mol. the corresponding free energy of formation is −237.2 kj/mol. answer in units of j/k.

The entropy change will be ΔS = - 150.0 J/K.

Explanation:

2 H₂(g) + O₂(g) → 2 H₂O(ℓ)  

ΔH = -258.8 kJ/mol

ΔG = - 237.2 kJ/mol

1 mol H₂ 2 g

x             3.66 g

x = 1.83 mol

1 mol H₂ 1 mol H₂O

1.83 mol H₂ 1.83 mol H₂O

ΔH = 1.83 mol x -258.8 kJ/mol

ΔH = - 478.78 kJ

ΔG = 1.83 mol x  - 237.2 kJ/mol

ΔG = 434.076 kJ

ΔG = ΔH - TΔS

ΔS = (ΔH - ΔG) / T

ΔS = [ - 478.78 kJ - (- 434.076 kJ)]/298 K

ΔS = - 150.0 J/K

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.