Calculate entropy change ( S ) and free energy change ( G ) for the conversion of ice into water at equilibrium when enthalpy change ( H ) is 9\ KJ/mol .

At equilibrium, the Gibbs free energy change ( G ) for any process is zero. This is because the system is in a state of equilibrium where the forward and reverse processes occur at equal rates, leading to no net change in free energy. Given: Enthalpy change ( H ) = 9\ kJ/mol Temperature ( T ) = 273\ K (melting point of ice) For the conversion of ice to water at equilibrium: G = H T S = 0 Rearranging to solve for entropy change ( S ): S = HT Substitute the given values: S = 9\ kJ/mol273\ K = 0.033\ kJ/(mol K) = 33\ J/(mol K) Thus, the entropy change ( S ) is 33 J/(mol·K) and the free energy change ( G ) is 0 kJ/mol .

Calculate entropy change ( S ) and free energy change ( G ) for the conversion of ice… – Chemistry NEB 2076 (old course) Solution

At equilibrium, the Gibbs free energy change ( $Δ G$ ) for any process is zero. This is because the system is in a state of equilibrium where the forward and reverse processes occur at equal rates, leading to no net change in free energy.

Given:

Enthalpy change ( $Δ H$ ) = $9 kJ/mol$
Temperature ( $T$ ) = $273 K$ (melting point of ice)

For the conversion of ice to water at equilibrium: $Δ G = Δ H - T Δ S = 0$

Rearranging to solve for entropy change ( $Δ S$ ): $Δ S = \frac{Δ H}{T}$

Substitute the given values: $Δ S = \frac{9 kJ/mol}{273 K} = 0.033 kJ/ (mol \cdot K) = 33 J/ (mol \cdot K)$

Thus, the entropy change ( $Δ S$ ) is 33 J/(mol·K) and the free energy change ( $Δ G$ ) is 0 kJ/mol.

Calculate entropy change ( S ) and free energy change ( G ) for the conversion of ice into water at equilibrium when enthalpy change ( H ) is 9\ KJ/mol .

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