Calculate a) ∆G° and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298K NO (g) + ½ O2

Exercises · 6.24

Q. Calculate a) ∆G° and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298K NO

(g) + ½ O2

(g) ⇌ NO2

(g) where ∆fG° (NO2) = 52.0 kJ/mol ∆fG° (NO) = 87.0 kJ/mol ∆fG° (O2) = 0 kJ/mol

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Concept understanding — Gibbs Free Energy from K

Gibbs Free Energy from $K$ — The Bridge Between Thermodynamics and Equilibrium

Imagine you're pushing a heavy box across a rough floor. You push hard, but the box barely moves. The potential to move is there — you're applying force — but the actual motion is tiny. That's the difference between thermodynamic spontaneity (the push) and equilibrium (where the box sits, barely budging).

Gibbs Free Energy ( $\Delta G$ ) tells you the push — whether a reaction can happen. The equilibrium constant $K$ tells you how far it actually goes before stopping. The equation that links them is one of the most powerful in chemistry:

\Delta G^\circ = -RT \ln K

Let's unpack this from the ground up.

Step 1: What is $\Delta G$ ?

Gibbs Free Energy change ( $\Delta G$ ) measures the maximum useful work a reaction can do at constant temperature and pressure. More practically:

If $\Delta G < 0$ : the reaction is spontaneous (it can happen on its own).
If $\Delta G > 0$ : the reaction is non-spontaneous (it needs energy input).
If $\Delta G = 0$ : the system is at equilibrium — no net change.

But here's the catch: $\Delta G$ depends on how much reactant and product you have at any moment. It's not a fixed number.

Step 2: Standard vs. Non-standard Conditions

Chemists define a standard state (pure substances at 1 bar, 1 M concentration for solutions, 25°C usually). Under those conditions, the free energy change is called $\Delta G^\circ$ (standard Gibbs free energy change).

But real reactions rarely start at standard conditions. So we have:

\Delta G = \Delta G^\circ + RT \ln Q

where $Q$ is the reaction quotient (ratio of products to reactants at that instant, raised to their stoichiometric coefficients).

Note

$R$ is the gas constant (8.314 J/mol·K), $T$ is temperature in Kelvin. The $\ln$ is natural log.

Step 3: At Equilibrium — The Key Insight

At equilibrium, the reaction has no net tendency to go forward or backward. That means:

\Delta G = 0

And the reaction quotient $Q$ becomes exactly the equilibrium constant $K$ .

So plug into the equation:

0 = \Delta G^\circ + RT \ln K

Rearrange:

\Delta G^\circ = -RT \ln K

That's it. This single equation connects a thermodynamic property ( $\Delta G^\circ$ ) with a concentration-based constant ( $K$ ).

Step 4: What This Tells You

$\Delta G^\circ$ value	$K$ value	Meaning
Negative ( $<0$ )	$K > 1$	Products favoured at equilibrium
Zero ( $=0$ )	$K = 1$	Equal amounts at equilibrium
Positive ( $>0$ )	$K < 1$	Reactants favoured at equilibrium

Watch out

A negative $\Delta G^\circ$ does not mean the reaction is fast — only that it's thermodynamically favourable. Kinetics (activation energy) is a separate story.

Step 5: A Concrete Example

Consider the reaction: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$

At 25°C, $\Delta G^\circ = -33.3 \text{ kJ/mol}$ . Using $R = 8.314 \text{ J/mol·K}$ :

$-33,300 = -(8.314)(298) \ln K$ …

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The standard Gibbs free energy change for a reaction equals the sum of formation energies of products minus reactants. Here $\Delta G^\circ = -35.0 \text{ kJ/mol}$ , which gives an equilibrium constant $K_p = 1.37 \times 10^6$ at 298 K.

Why this approach works

Gibbs free energy tells us whether a reaction is spontaneous and how far it will proceed toward products at equilibrium. The standard Gibbs free energy of reaction, $\Delta G^\circ$ , is calculated from tabulated formation values using Hess's law: the energy change for the overall reaction is the difference between the energy required to form products and that to form reactants from their elements.

Once we have $\Delta G^\circ$ , the fundamental relationship between thermodynamics and equilibrium connects it to the equilibrium constant through:

\Delta G^\circ = -RT \ln K

This equation captures a beautiful idea: a negative $\Delta G^\circ$ (spontaneous forward reaction) corresponds to $K > 1$ (products favored), while a positive value means $K < 1$ (reactants favored).

Solution

Part (a): Calculate $\Delta G^\circ$

Write the formation energy equation For any reaction, the standard Gibbs free energy change is:

\Delta G^\circ_{\text{rxn}} = \sum \Delta_f G^\circ(\text{products}) - \sum \Delta_f G^\circ(\text{reactants})

Each term is multiplied by its stoichiometric coefficient.

Identify the stoichiometry Our reaction is:

\text{NO}(g) + \tfrac{1}{2}\text{O}_2(g) \rightleftharpoons \text{NO}_2(g)

Products: 1 mol NO₂

Reactants: 1 mol NO, $\tfrac{1}{2}$ mol O₂

Substitute the given values

\Delta G^\circ = [1 \times \Delta_f G^\circ(\text{NO}_2)] - [1 \times \Delta_f G^\circ(\text{NO}) + \tfrac{1}{2} \times \Delta_f G^\circ(\text{O}_2)]

\Delta G^\circ = [1 \times 52.0] - [1 \times 87.0 + \tfrac{1}{2} \times 0]

\Delta G^\circ = 52.0 - 87.0 = -35.0 \text{ kJ/mol}

The negative value tells us the forward reaction is thermodynamically favorable under standard conditions.

Note

Elemental oxygen in its standard state (O₂ gas) has $\Delta_f G^\circ = 0$ by definition, since formation energies measure the energy to form a substance from its elements.

Part (b): Calculate the equilibrium constant

Use the Gibbs-equilibrium relationship Starting from:

\Delta G^\circ = -RT \ln K

Rearrange to solve for $K$ :

\ln K = -\frac{\Delta G^\circ}{RT}

K = e^{-\Delta G^\circ / RT}

Convert units and substitute …

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Q.Calculate a) ∆G° and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298K NO

Why this approach works

Solution

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