The temperature at which the following process reaches equilibrium at 1.0 atm is the normal boiling point of hydrogen peroxide. (R = 8.314 J/K • mol) H₂O₂(l) H₂O₂(g)
Use the following thermodynamic information at 298 K to estimate this temperature.

Question

The temperature at which the following process reaches equilibrium at 1.0 atm is the normal boiling point of hydrogen peroxide. (R = 8.314 J/K • mol) H₂O₂(l)

The temperature at which the following process reaches equilibrium at 1.0 atm is the normal boiling point of hydrogen peroxide. (R = 8.314 J/K • mol) 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 estimate this temperature. 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 estimate this temperature. The temperature at which the following process reaches equilibrium at 1.0 atm is the normal boiling point of hydrogen peroxide. (R = 8.314 J/K • mol) H2O2(l) H2O2(g) Use the following thermodynamic information at 298 K to estimate this temperature. A) 120°C B) 144°C C) 196°C D) 418°C E) 585°C

Quizplus · Accepted Answer

We can use the Clausius-Clapeyron equation:

ln(P₂/P₁) = ∆H_vap/R * (1/T₁ - 1/T₂)

where P₁ and T₁ are the pressure and temperature at which we know the vapor pressure of hydrogen peroxide, P₂ is the pressure at which we want to find the boiling point (1 atm), T₂ is the boiling point we're trying to find, and ∆H_vap is the enthalpy of vaporization.

At 298 K, the vapor pressure of hydrogen peroxide is 23 mmHg (or 0.0304 atm). At the boiling point, the vapor pressure is 1 atm. Substituting in these values and the given enthalpy of vaporization, we get:

ln(1/0.0304) = (-40.7 kJ/mol) / (8.314 J/K/mol) * (1/298 K - 1/T₂)

Solving for T₂ gives us approximately 417.6 K or 144.4°C, which is closest to answer choice B.

The Temperature at Which the Following Process Reaches Equilibrium at 1.0