2025 MCAT General Chemistry Practice Exam – Your Comprehensive All-in-One Study Guide for Success!

Question: 1 / 400

What happens when K is less than 1 in terms of ∆G°?

∆G° is less than 0

∆G° is equal to 0

∆G° is greater than 0

When the equilibrium constant \( K \) is less than 1, it indicates that the products of the reaction are less favored compared to the reactants at equilibrium. In thermodynamic terms, this is associated with a positive standard Gibbs free energy change (\( \Delta G^\circ \)) for the corresponding reaction.

The relationship between the equilibrium constant and the standard Gibbs free energy change is given by the equation:

\[

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

\]

Here, \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( K \) is the equilibrium constant. When \( K < 1 \), the natural logarithm of \( K \) becomes negative (\( \ln K < 0 \)). Substituting this negative value into the equation leads to:

\[

\Delta G^\circ = -RT \cdot (\text{a negative value}) = \text{a positive value}

\]

As a result, \( \Delta G^\circ \) is greater than 0 when \( K < 1 \), confirming that the reaction is non-spontaneous under standard conditions.

Understanding the implications of different values of \( K \)

Get further explanation with Examzify DeepDiveBeta

Does not apply

Next Question

Report this question

Download 2025 MCAT General Chemistry Practice Exam – Your Comprehensive All-in-One Study Guide for Success! PDF Study Guide
Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy