Calculate the value of the ∆S˚rxn for each given reactions:

Question

1. 3NO(g) → N2O(g) + NO2(g)

2. 3H2(g) + Fe2O3(s) → 2Fe(s) + 3H2O(g)

3. P4(s) + 5O2(g) → P4O10(s)

4. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g)

Answer 1

. The K constant from values looked up when .

3NO ==> NO2 + N2O.

Is 6.68x 10^8

This is from 3NO: 0.0015, N2O: 1.25, NO2: 1.80

At equilibrium.

If you use this equation

∆G° = -RTlnK

And the correct R constant 8.314 J/K*mol

You can find ∆G°

For .1) evaluating the equation:

∆G° = -8.314 J/K*mol298Kln(6.68x10^8)

∆G° =-8.314J/K*mol *298K-18.82

∆G° = 46,638 J/mol. or 50,434J/mol based on rounding

Then use the second equation:

∆G° = ∆H° -T∆S°.

46,638 = X- 298∆S°

X will be you looking up the standard enthalpy connected with this reaction or ∆H°:

You can look up values and use Hess's Law to find out what this value would be:

From:

-81.6 kJ/mol + 56.55kJ/mol = -25.05 kJ/mol

Now go back and look at how the reaction is written here in the problem and make it reflect:

Here it's written:

3NO==> N2O. + NO2.

With the enthalpy reaction using Hess's law it is written:

N2O + NO2 ==> 3NO

So you must flip the energy sign and you will have:

25.05 kJ/mol

∆H° = 25.05kJ/mol

Now express this as Joules so you will have uniform units.

1000 J = 1kJ.

25.05kJ/mol x 1000 J/1kJ. kJ cancel top to bottom and you have Joules/mol

And solve for ∆S°

∆G° = ∆H° -T∆S°.

And now solve for ∆S°

46,638 J/mol= 25,050J/mol -298K∆S°

21,588J/mol= (-298K∆S°)/-298K

∆S° = -72.44J/mol

If entropy is positive there is more disorder, but if it is negative the opposite is the case. There is less disorder connected with this reaction. It's not quite a spontaneous reaction.

This also lines up with a positive ∆H°. If it's positive, it is endothermic and therefore requires energy for the reaction to take place.

You can follow this process with parts 2-4 of this problem.