Question

At \( 298 \) K, \( \Delta U^\circ \) and \( \Delta S^\circ \) for the following reaction are \( -10.5 \) kJ and \( +44.1 \) J\( K^{-1} \); \( 2X(g) + Y(g) \longrightarrow 2Z(g) \). What is \( \Delta G^\circ \) (in kJ) for this reaction? (\( R = 8.314 \) J\( K^{-1} \)mol\(^{-1} \))

• A. \( +0.164 \)
• B. \( -26.119 \)
• C. \( -2.6119 \)
• D. \( -0.082 \)

Hint:

For Gibbs free energy calculations, use: \[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \] where \( T \) is in Kelvin and \( \Delta S^\circ \) is converted to kJ before computation.

Answer 1

The Correct Option is B

Step 1: Gibbs Free Energy Formula The Gibbs free energy change is given by: \[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \] Since \( \Delta H^\circ \) and \( \Delta U^\circ \) are related by: \[ \Delta H^\circ = \Delta U^\circ + \Delta n_g RT \] Here, \( \Delta n_g = 2 - (2 + 1) = -1 \). Thus, \[ \Delta H^\circ = -10.5 + (-1 \times 8.314 \times 298 \times 10^{-3}) \] \[ = -10.5 - 2.477 \approx -12.977 \text{ kJ} \] Step 2: Computing \( \Delta G^\circ \) \[ \Delta G^\circ = -12.977 - (298 \times 44.1 \times 10^{-3}) \] \[ = -12.977 - 13.142 \] \[ = -26.119 \text{ kJ} \] Conclusion Thus, the correct answer is: \[ -26.119 \text{ kJ} \]