At 298 K, Kp for the reaction N2O4(g) ⇌ 2NO2(g) is 0.98. Predict whether the reaction is spontaneous or not.

NCERT Exemplar · Q30

Q. At 298 K, Kp for the reaction N2O4(g) ⇌ 2NO2(g) is 0.98. Predict whether the reaction is spontaneous or not.

Rajasthan RbseShort· 2mImportance★★★★★est

67% · 66/98 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 $R = 8.314 \text{ J/mol·K}$ :

$-33,300 = -(8.314)(298) \ln K$ …

Start your 14-day free trial to unlock the full solution →

The spontaneity of a reaction is determined by the sign of $\Delta G^\circ$ (standard Gibbs free energy change), which is related to the equilibrium constant $K_p$ by $\Delta G^\circ = -RT \ln K_p$ . Here, $K_p = 0.98 < 1$ , so $\ln K_p$ is negative, making $\Delta G^\circ$ positive — the reaction is non-spontaneous under standard conditions at 298 K.

The question asks whether the reaction $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ is spontaneous at 298 K, given $K_p = 0.98$ . The key is to connect the equilibrium constant to Gibbs free energy.

Spontaneity under standard conditions (1 bar pressure for gases, 1 M concentration for solutions, pure solids/liquids) is governed by the sign of $\Delta G^\circ$ . A negative $\Delta G^\circ$ means the forward reaction is spontaneous; a positive $\Delta G^\circ$ means it is non-spontaneous (the reverse reaction is spontaneous). The relationship is:

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

Here, $R$ is the gas constant (8.314 J/mol·K), $T$ is the temperature in Kelvin, and $K_p$ is the equilibrium constant in terms of partial pressures. This formula comes from the thermodynamic condition for equilibrium: at equilibrium, $\Delta G = 0$ , and $\Delta G = \Delta G^\circ + RT \ln Q$ , where $Q$ is the reaction quotient. Setting $Q = K_p$ at equilibrium gives the equation above.

Now, let's work through the calculation step by step.

Identify the given values.

$T = 298 \text{ K}$ , $K_p = 0.98$ , $R = 8.314 \text{ J mol}^{-1} \text{K}^{-1}$ .
Compute $\ln K_p$ .

Since $K_p = 0.98$ , which is slightly less than 1, $\ln(0.98)$ is a small negative number.

\ln(0.98) \approx -0.0202

(You can verify: $e^{-0.0202} \approx 0.98$ .)

Plug into the formula.

\Delta G^\circ = - (8.314)(298) \ln(0.98)

First compute $RT$ : $8.314 \times 298 \approx 2477.6 \text{ J/mol}$ .

Then:

\Delta G^\circ = -2477.6 \times (-0.0202) = +50.0 \text{ J/mol}

More precisely, $2477.6 \times 0.0202 \approx 50.05$ , so $\Delta G^\circ \approx +50 \text{ J/mol}$ .

Interpret the sign. $\Delta G^\circ$ is positive (about $+50 \text{ J/mol}$ ). Under standard conditions, a positive $\Delta G^\circ$ means the forward reaction is non-spontaneous. The reverse reaction (2 NO₂ → N₂O₄) would be spontaneous. …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ29 — Increase in enthalpy of the surroundings is equal to decrease in entha…→Next questionQ31 — A sample of 1.0 mol of a monoatomic ideal gas is taken through a cycli…

Q.At 298 K, Kp for the reaction N2O4(g) ⇌ 2NO2(g) is 0.98. Predict whether the reaction is spontaneous or not.

🎓Unlock everything free for 14 days

Unlock everything free for 14 days