Question

Why is the equilibrium constant (K) temperature-dependent?

• A. Because $\Delta G^\circ$ depends on T
• B. Because pressure affects K
• C. Because entropy is temperature-independent
• D. Because fugacity coefficients change with T

Hint:

Remember: $K$ is linked to $\Delta G^\circ$ via $\Delta G^\circ = -RT \ln K$. Since $\Delta G^\circ$ depends on temperature, so does $K$.

Answer 1

The Correct Option is A

The relationship between the standard Gibbs free energy change and the equilibrium constant $K$ is given by the fundamental thermodynamic equation:
\[ \Delta G^\circ = -RT \ln K \]
Where:
$R$ = universal gas constant
$T$ = absolute temperature in Kelvin
$K$ = equilibrium constant
Rearranging gives:
\[ \ln K = -\frac{\Delta G^\circ}{RT} \]
From this equation, we see that the value of $K$ depends on both temperature $T$ and the value of $\Delta G^\circ$. But $\Delta G^\circ$ itself is not a constant—it is a function of temperature.
Recall that:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]
Since both enthalpy $\Delta H^\circ$ and entropy $\Delta S^\circ$ can be temperature-dependent, $\Delta G^\circ$ also changes with temperature. As a result, the equilibrium constant $K$ becomes temperature-dependent.
Now let's review the other options:
- (2) Pressure may affect the position of equilibrium but does not define the fundamental temperature dependency of $K$.
- (3) Entropy is not temperature-independent—it typically changes with temperature, making this statement incorrect.
- (4) Fugacity coefficients are relevant in real gases but they do not primarily explain why $K$ is temperature-dependent in standard thermodynamics.
Hence, the most accurate reason for $K$ being temperature-dependent is because $\Delta G^\circ$ depends on $T$.