Consider the reaction cs2(l) = cs2(g). calculate h° and s° using data tables. estimate the boiling point of carbon disulfide, in °c, from this data. estimate the vapor pressure of cs2 at 20°c, in torr. (hint: what are the units of standard conditions? )

ΔH° reaction= 27.4 kJ/mol

ΔS° reaction= 106.8 J/mol.K

Boiling point= -30.64 °C

Pv = 0.079  torr

Explanation:

To solve this problem, we have to analyze the process that we are going to study:

Carbon disulfide is changing from liquid state to gas state.

As you must know, the gas state is more entropic than the liquid state, so we going to expect an increase in S, and also involves a change in heat, an increase in H.

a. Calculate ΔH°  and ΔS°

Remember that the ° means “standard conditions”, you can find this values for de CS2 in a table, for each one of  these state:

CS2 (l) :

ΔH° = 87,9 kJ/mol

ΔS° = 131,0 J/mol.K

CS2 (g) :

ΔH° = 115,3 kJ/mol

ΔS° =  237,8 J/mol.K

Note that the ΔH° of the product (CS2 (g)) is bigger than reactant CS2 (l).

With this data, you can calculate the ΔH° for the reactions using the following equation:

ΔH° reaction= ΔH° products - ΔH° reactants

So:

ΔH° reaction= 115,3 kJ/mol-87,9 kJ/mol = 27.4 kJ/mol

And as is obtained a ΔH° reaction > 0 , iit means positive, it is an endothermic process, so, these reactions require to apply external energy or heat to be doing.

Now we going to calculate ΔS°reaction

ΔS° reaction= ΔS° products - ΔS° reactants

Where

ΔS° products = 237,8 J/mol.K

ΔS° reactants =  131,0 J/mol.K

Replacing on the equation:

ΔS° reaction= 237,8 J/mol.K - 131,0 J/mol.K

ΔS° reaction= 106.8 J/mol.K

b. Estimate the boiling point of carbon disulfide, in °C, from this data.

To make this estimation, we have to assume constant pressure:

To calculate this boiling temperature, we have to use the following equation who relates entropy with enthalpy and free Gibbs energy:

ΔG° = ΔH°   – T ΔS°

The values of  ΔG°  can be found in the same table of H°  and  ΔS°:

CS2 (l) :

ΔG° = 63,6KJ/mol

CS2 (g) :

ΔG° =65,1KJ/mol

With this we can calculate ΔG°:

ΔG°= ΔG° products - ΔG° reactants

ΔG°= 65,1kJ/mol -63,6 kJ/mol = 1.5  kJ/mol

And we this value we replace in the equation as follow:

ΔG° = ΔH°   – T ΔS°

1.5  kJ/mol = 27.4  kJ/mol – T ×106.8 J/mol.K

And from this equation, obtain T:

-106.8 J/mol.K × T =  1.5  kJ/mol – 27.4  kJ/mol

-106.8 J/mol.K × T = -25.9 kJ/mol

aquifers are bounded above and below by impermeable rock. the driving force of groundwater movement in confined aquifers is pressure, rather than gravity.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3