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For the process: [tex]A(g) \rightarrow B(g)[/tex], given [tex]\Delta H^\circ = 159 \text{ kJ}[/tex] and [tex]\Delta S^\circ = 159 \text{ J/K}[/tex], calculate [tex]\Delta G^\circ[/tex].

Is the process spontaneous at 0°C and 1 atm pressure?

Answer :

Final answer:

To determine if the process A(g) → B(g) is spontaneous at 0°C and 1 atm pressure, we can calculate the standard Gibbs free energy change (ΔG°) using the equation ΔG° = ΔH° - TΔS°. If ΔG° is negative, the process is spontaneous.

Explanation:

To calculate the standard Gibbs free energy change (ΔG°) for the process A(g) → B(g), we can use the equation ΔG° = ΔH° - TΔS°, where ΔH° is the standard enthalpy change and ΔS° is the standard entropy change. In this case, ΔH° = 159 kJ and ΔS° = 159 J/K. To determine if the process is spontaneous at 0°C and 1 atm pressure, we need to consider the sign of ΔG°.


At constant temperature, ΔG° can be determined using the equation ΔG° = ΔH° - TΔS°. If ΔG° is negative, the process is spontaneous. If ΔG° is positive, the process is non-spontaneous. If ΔG° is zero, the process is at equilibrium.

Substituting the given values, we have ΔG° = 159 kJ - (273 + 0) K * (159 J/K) = -159 kJ. Since ΔG° is negative, the process is spontaneous at 0°C and 1 atm pressure.

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