The temperature at which the following process reaches...

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. 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

Transcript

00:01 So the first what we have is the temperature at which the process will reach the equilibrium at one atmosphere that is the normal boiling point of the hydrogen peroxide.

00:16 For that here we have the equation that we can represent it like this and we are given with the several values for both of these substances.

00:25 So your first value that we have is delta hf0 kilo -zoules per mole, delta g -0 that is also in the kilozoules per mole, and the s -not values, that is the zholes per kelvin mole.

00:41 So here what we have minus 187 .7, minus 120 .3, 109 .6, minus 136 .3, minus 105 .6, and 2 .306 .6.

00:55 32 .7.

00:57 Now we have one relation that is delta j is equal to delta h minus t delta s not.

01:05 We need to find out this t term.

01:08 Here, the change in gives energy.

01:11 This corresponds to zero.

01:13 When the compound is undergoing the vaporization, when the compound is undergoing vaporization at the normal boiling point.

01:22 So here it is also going at normal boiling point.

01:24 So that implies.

01:25 The gives free energy of vapor will be is equal to the gives free energy of the liquid so here it will be the zero so from here what we get is delta h is equal to t delta s not so from here what we have is t is equal to delta h upon delta s not so from here we can find out these both values because we are given with that so delta h will be is equal to for the gaseous minus liquid so for the gases it is minus 136 .3 minus and then here we are minus 187 .7...

Question

Please give Ace some feedback