The equilibrium constant for a reaction is 10. What will be the value of $\Delta G^\ominus$? $R = 8.314$ JK$^{-1}$ mol$^{-1}$, $T = 300$ K.

The equilibrium constant for a reaction is 10. What will be the value of Δ Gominus? R = 8.314 JK⁻¹ mol⁻¹, T =

Exercises · 5.20

\Delta G^\ominus

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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 $\Delta G^\ominus$ is directly related to the equilibrium constant $K$ by $\Delta G^\ominus = -RT \ln K$ . For $K = 10$ , $R = 8.314$ J K⁻¹ mol⁻¹, and $T = 300$ K, the value is approximately –5744 J mol⁻¹ (or –5.744 kJ mol⁻¹).

The connection between $\Delta G^\ominus$ and $K$ is one of the most elegant results in chemical thermodynamics. It tells us that the standard free energy change — the maximum useful work obtainable from a reaction under standard conditions — is simply a logarithmic function of the equilibrium constant. Why? Because at equilibrium, the reaction quotient $Q$ equals $K$ , and the Gibbs free energy change $\Delta G$ is zero. The equation $\Delta G = \Delta G^\ominus + RT \ln Q$ then gives $\Delta G^\ominus = -RT \ln K$ .

This means: if $K > 1$ , $\ln K$ is positive, so $\Delta G^\ominus$ is negative — the reaction is spontaneous in the forward direction under standard conditions. Here $K = 10$ , so we expect a negative value.

Let’s compute it step by step.

Write the fundamental relation The formula is:

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

where $R$ is the gas constant, $T$ is the absolute temperature, and $K$ is the equilibrium constant (dimensionless when using standard states).

Plug in the given values $R = 8.314\ \text{J K}^{-1} \text{mol}^{-1}$ , $T = 300\ \text{K}$ , $K = 10$ . So:

\Delta G^\ominus = - (8.314) \times (300) \times \ln(10)

Evaluate $\ln(10)$

$\ln(10) \approx 2.302585$ . (A useful constant to remember: $\ln(10) \approx 2.303$ .)
Multiply step by step

First, $RT = 8.314 \times 300 = 2494.2\ \text{J mol}^{-1}$ .

Then, $RT \ln K = 2494.2 \times 2.302585 \approx 5744.0\ \text{J mol}^{-1}$ . …

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Q.The equilibrium constant for a reaction is 10. What will be the value of ΔG⊖\Delta G^\ominusΔG⊖? R=8.314R = 8.314R=8.314 JK−1^{-1}−1 mol−1^{-1}−1, T=300T = 300T=300 K.

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Q.The equilibrium constant for a reaction is 10. What will be the value of $\Delta G^\ominus$ ? $R = 8.314$ JK $^{-1}$ mol $^{-1}$ , $T = 300$ K.

Unlock everything free for 14 days