The relationship between the standard free energy change (ΔG⊖) and the equilibrium constant (K) is given by the equation:

ΔG⊖ = -RT ln K

where R is the gas constant and T is the temperature in Kelvin.

If ΔG⊖ 0, this means that the reaction is not spontaneous in the forward direction under standard conditions.

The natural logarithm (ln) of a number between 0 and 1 is negative. Therefore, if ΔG⊖ 0, then ln K must be negative.

This implies that K must be less than 1.

So, for the reaction H2(g) + I2(g) ⇔ 2HI (g), if the standard free energy is ΔG⊖ 0, the equilibrium constant (K) would be less than 1.

Question

The relationship between the standard free energy change (ΔG⊖) and the equilibrium constant (K) is given by the equation:

ΔG⊖ = -RT ln K

where R is the gas constant and T is the temperature in Kelvin.

If ΔG⊖ > 0, this means that the reaction is not spontaneous in the forward direction under standard conditions.

The natural logarithm (ln) of a number between 0 and 1 is negative. Therefore, if ΔG⊖ > 0, then ln K must be negative.

This implies that K must be less than 1.

So, for the reaction H2(g) + I2(g) ⇔ 2HI (g), if the standard free energy is ΔG⊖ > 0, the equilibrium constant (K) would be less than 1.

Knowee AI · Accepted Answer

The relationship between the standard free energy change (ΔG⊖) and the equilibrium constant (K) is given by the equation:

ΔG⊖ = -RT ln K

where R is the gas constant and T is the temperature in Kelvin.

If ΔG⊖ > 0, this means that the reaction is not spontaneous in the forward direction under standard conditions.

The natural logarithm (ln) of a number between 0 and 1 is negative. Therefore, if ΔG⊖ > 0, then ln K must be negative.

This implies that K must be less than 1.

So, for the reaction H2(g) + I2(g) ⇔ 2HI (g), if the standard free energy is ΔG⊖ > 0, the equilibrium constant (K) would be less than 1.

For the reaction H2(g) + I2(g) ⇔ 2HI (g), the standard free energy is ΔG⊖ > 0. The equilibrium constant (K ) would be