Calculate Free energy change (delta G) for the following reactions at 25°C:

Calculate ΔG^° for the following reactions at 25 °C:

(a) \(N_2(g) + O_2(g)\rightarrow 2NO (g)\)

(b)\(H_2O(l) \rightarrow H_2O (g)\)

(c) \(2C_2H_2(g)+ 5O_2(g) \rightarrow 4CO_2(g) + 2H_2O (l)\)

Standard free energy formation value :

Substance	Standard free energy formation values ( in kJ/mol)
N2 (g)	0
O2 (g)	0
NO (g)	86.7
H2O (l)	-237.2
H2O (g)	-228.6
C2 H2(g)	209.2
CO2(g)	-394.4

Interpretation

The quantity we need is the standard Gibbs free energy change of reaction, denoted by \(\Delta G^\circ_{\text{rxn}}\).

Gibbs free energy combines two competing tendencies:

• Enthalpy (\(H\)): systems tend to move toward lower energy.
• Entropy (\(S\)): systems tend to move toward greater disorder.

The relationship is:\(G = H – TS\)

For a chemical reaction under standard conditions,\[
\Delta G^\circ
=
\sum n\,\Delta G_f^\circ(\text{products})
–
\sum n\,\Delta G_f^\circ(\text{reactants})
\]

where:

• \(\Delta G_f^\circ\) = standard Gibbs free energy of formation
• \(n\) = stoichiometric coefficient

A very important fact is that the standard free energy of formation of an element in its most stable standard state is zero. Thus,\(\Delta G_f^\circ(\mathrm{N_2(g)})=0\)\(\Delta G_f^\circ(\mathrm{O_2(g)})=0\)

---

Standard Gibbs Free Energies of Formation at 25 °C

We use the commonly tabulated values:

Substance	\(\Delta G_f^\circ\) (kJ mol⁻¹)
\(\mathrm{N_2(g)}\)	0
\(\mathrm{O_2(g)}\)	0
\(\mathrm{NO(g)}\)	+86.6
\(\mathrm{H_2O(l)}\)	−237.1
\(\mathrm{H_2O(g)}\)	−228.6
\(\mathrm{CO_2(g)}\)	−394.4
\(\mathrm{C_2H_2(g)}\)	+209.2

---

(a) \(\mathrm{N_2(g)+O_2(g)\rightarrow 2NO(g)}\)

Concept Application

Both reactants elements are in their standard states, so their formation free energies are zero.

Solution

Using \[
\Delta G^\circ_{\text{rxn}}
=
\sum n\,\Delta G_f^\circ(\text{products})
–
\sum n\,\Delta G_f^\circ(\text{reactants})
\]

$$\Delta G^\circ =2(\Delta G_f^\circ(\mathrm{NO})) -[0+0]$$$$=2(86.6)$$$$=173.2\ \text{kJ mol}^{-1}$$

Answer:$$\boxed{\Delta G^\circ = +173.2\ \text{kJ mol}^{-1}}$$

Hence, the positive value indicates that formation of NO from nitrogen and oxygen is non-spontaneous under standard conditions.

---

(b) \(\mathrm{H_2O(l)\rightarrow H_2O(g)}\)

Concept Application

This is simply vaporization of water. The chemical identity remains the same, but the physical state changes.

Solution

$$\Delta G^\circ =\Delta G_f^\circ(\mathrm{H_2O(g)}) -\Delta G_f^\circ(\mathrm{H_2O(l)})$$$$=(-228.6)-(-237.1)$$$$=+8.5\ \text{kJ mol}^{-1}$$

Answer:$$\boxed{\Delta G^\circ = +8.5\ \text{kJ mol}^{-1}}$$

The positive value means that at 25 °C and standard pressure, liquid water is more stable than water vapor.

---

(c) \(\mathrm{2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(l)}\)

Concept Application

This is the combustion of acetylene. Oxygen is an element in its standard state, so its formation free energy is zero.

Solution

First calculate the products:$$4(\Delta G_f^\circ(\mathrm{CO_2})) +2(\Delta G_f^\circ(\mathrm{H_2O(l)}))$$$$=4(-394.4)+2(-237.1)$$$$=-1577.6-474.2$$$$=-2051.8\ \text{kJ}$$

Now calculate the reactants:$$2(\Delta G_f^\circ(\mathrm{C_2H_2})) +5(\Delta G_f^\circ(\mathrm{O_2}))$$$$=2(209.2)+5(0)$$$$=418.4\ \text{kJ}$$

Therefore,$$\Delta G^\circ =(-2051.8)-(418.4)$$$$=-2470.2\ \text{kJ mol}^{-1}$$

Answer:$$\boxed{\Delta G^\circ = -2470.2\ \text{kJ mol}^{-1}}$$

The large negative value shows that acetylene combustion is strongly spontaneous under standard conditions.

---

Final Answers

$$\boxed{\mathrm{N_2(g)+O_2(g)\rightarrow 2NO(g)}}$$$$\boxed{\Delta G^\circ = +173.2\ \text{kJ mol}^{-1}}$$$$\boxed{\mathrm{H_2O(l)\rightarrow H_2O(g)}}$$$$\boxed{\Delta G^\circ = +8.5\ \text{kJ mol}^{-1}}$$$$\boxed{\mathrm{2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(l)}}$$$$\boxed{\Delta G^\circ = -2470.2\ \text{kJ mol}^{-1}}$$

Insight

A quick way to remember Gibbs free energy calculations is:

Products minus reactants, using formation values.

Also remember the special shortcut:

“Importantly, the standard free energy of formation of any element in its standard state is zero.”

That single fact often eliminates many terms and makes reaction free-energy calculations much easier.