Use the thermodynamic data table to calculate ∆Ssystem, ∆Ssorroundings, ∆Suniverse and G at 25°C.

Question

1. 3N2O4(g) → 2NO(g) + 2N2O(g) + 4°2(g)

2. 4NH3(g) 5O2(g) → 4NO(g) 6H2O(g)

Answer 1

Hi Candice, although we don't have a thermodynamics table to calculate these values. I can show you the general formulas to calculate these, and basing on my thermodynamic data table values I have, can give you a general idea how to solve each one. I will show only the first one, as the concept applies to any chemical equation.

1. 3N2O4(g) → 2NO(g) + 2N2O(g) + 4O2(g)

First we must note the thermodynamic values of each for both delta S(J/mol*K) and delta H(kJ/mol)

Delta S(J/mol*K)

N2O4(g) = 304

NO(g) = 210

N2O = 220

O2(g) = 205

Delta Hf(kJ/mol)

N2O4 = +9.2

NO(g) = +90.3

N2O(g) = +82.0

O2(g) = 0

Let's calculate Delta S system:

Products S = 2*210 + 2*220 + 4*205 = 1680 J/mol*K

Reactants S = 3*304 = 912J/mol*K

Delta S System = 1680 - 912 = 768 J/mol*K

** notice how we multiply the delta s values by the coefficients of the given reaction, for ex. 4 Oxygens multiply by 205J/mol*K

Delta H will be the same exact format

Delta H = 344.6 - 27.6 = +317.0 kJ

Delta S_surroundings would be using the equation: delta s surr = -delta H/T(Temperature) = -317000J/298K = -1064 J/mol

** note here the equation -delta H is found from the previous step before, and the temperature is the standard temperature at 298K or 25 Degrees Celsius + 273 = 298K, we also converted the delta H from kJ to J, 317.0 kJ -> 317000 J

Delta S_universe would be delta s universe = (s_surr) + (s_system) = 768 + (-1064) = -296J/mol*K

and finally Delta G, use equation delta G= deltaH - T*delta S system = 317 - 228.9 = +88.1kJ/mol

As you can the process can get very repetitive, but as long as you have standard equations, balanced chemical reactions, standard delta s and delta h data values, follow this format for any equation, and it can be calculated. Hope this helps, thank you!