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

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Multiple Choice

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

Explanation:
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 \( K \)

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 )

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