Calculate the entropy change when 4.31 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.

Explanation:

ΔG  =  Δ H - T ΔS

Given , ΔG = -237.2 kJ / mol

Δ H = - 285.8 kJ / mol

T = 298

Putting the values in the equation

-237.2 = - 285.8 - 298 x ΔS

ΔS  = .163 kJ / mole K

moles of water formed = 4.31 / 2 = 2.155

entropy change required = 2.155 x .163 kJ / K

= 351.26 J / K .

stupid i wont let you get a chance

explanation: