Question

The temperature at which the following process reaches equilibrium at 1.0 atm is the normal boiling point of hydrogen peroxide. $H_2O_2(l) \rightleftharpoons H_2O_2(g)$ Use the following thermodynamic information at 298 K to determine this temperature. Substance: $\Delta H^\circ_f$ (kJ/mol): $\Delta G^\circ_f$ (kJ/mol): $S^\circ$ (J/K\cdot mol): 144°C 418°C 196°C 585°C 120°C $H_2O_2(l)$ $H_2O_2(g)$ -187.7 -136.3 -120.4 -105.6 109.6 232.7

          The temperature at which the following process reaches
equilibrium at 1.0 atm is the normal boiling point of
hydrogen peroxide.
$H_2O_2(l) \rightleftharpoons H_2O_2(g)$
Use the following thermodynamic information at 298 K to determine this
temperature.
Substance:
$\Delta H^\circ_f$ (kJ/mol):
$\Delta G^\circ_f$ (kJ/mol):
$S^\circ$ (J/K\cdot mol):
144°C
418°C
196°C
585°C
120°C
$H_2O_2(l)$  $H_2O_2(g)$
-187.7     -136.3
-120.4     -105.6
109.6      232.7

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:
Δ H^ (kJ/mol):
Δ G^ (kJ/mol):
S^∘ (J/K·mol):
144°C
418°C
196°C
585°C
120°C
H2O2(l) H2O2(g)
-187.7 -136.3
-120.4 -105.6
109.6 232.7

Added by Alicia S.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Chittaranjan Sahoo

Step 1

6 - (-120.4) = 14.8 kJ/mol ΔS = ΔS(products) - ΔS(reactants) = 232.7 - 109.6 = 123.1 J/mol K At equilibrium, ΔG = 0, so we can use the equation ΔG = ΔH - TΔS to find the temperature at which the process reaches equilibrium. We first need to find the change in Show more…

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The temperature at which the following process reaches equilibrium at 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: Î”Hf (kJ/mol): Î”G' (kJ/mol): Î”S (J/K mol): H2O2(l) -187.7 -120.4 109.6 H2O2(g) -136.3 -105.6 232.7