The temperature at which the following process reaches equilibrium at 1.0 atm is the normal boiling point of hydrogen peroxide. H₂O₂(l) H₂O₂(g)
Use the following thermodynamic information at 298 K to determine this temperature.
Substance: H₂O₂(l) H₂O₂(g)
ΔH°_f (kJ/mol) : -187.7 -136.3
ΔG°_f (kJ/mol) : -120.4 -105.6
S°(J/K·mol) : 109.6 232.7

Question

The temperature at which the following process reaches equilibrium at 1.0 atm is the normal boiling point of hydrogen peroxide. H₂O₂(l)

The temperature at which the following process reaches equilibrium at 1.0 atm is the normal boiling point of hydrogen peroxide. H<sub>2</sub>O<sub>2</sub>(l) H<sub>2</sub>O<sub>2</sub>(g) Use the following thermodynamic information at 298 K to determine this temperature. Substance: H<sub>2</sub>O<sub>2</sub>(l) H<sub>2</sub>O<sub>2</sub>(g) ΔH°<sub>f</sub> (kJ/mol) : -187.7 -136.3 ΔG°<sub>f</sub> (kJ/mol) : -120.4 -105.6 S°(J/K·mol) : 109.6 232.7 A) 120°C B) 144°C C) 196°C D) 418°C E) 585°C

H₂O₂(g)
Use the following thermodynamic information at 298 K to determine this temperature.
Substance: H₂O₂(l) H₂O₂(g)
ΔH°_f (kJ/mol) : -187.7 -136.3
ΔG°_f (kJ/mol) : -120.4 -105.6
S°(J/K·mol) : 109.6 232.7

Quizplus · Accepted Answer

To determine the temperature at which the given process reaches equilibrium, we need to use the equation ΔG° = -RTln(K), where K is the equilibrium constant and R is the gas constant. At the boiling point, the liquid and gas phases are in equilibrium, so we can use the equation:

ΔG°(boiling) = 0 = ΔG°(gas) - ΔG°(liquid)

Solving for T, we get:

T = ΔH°(gas) - ΔH°(liquid) / ΔS°(gas) - ΔS°(liquid) x R

Plugging in the values given:

T = (-136.3 - (-187.7)) / ((232.7 - 109.6) / 1000) x 8.314

T = 144°C

Therefore, the correct answer is B.

The Temperature at Which the Following Process Reaches Equilibrium at 1.0