Question

The $\Delta G^\circ$ for the reaction $3X(g)+2Y(g)\rightarrow3Z(g)$ at $298K$ is ($\Delta H^\circ=-13$ kJ mol$^{-1}$ and $\Delta S^\circ=-45$ J K$^{-1}$ mol$^{-1}$)

• A. $-0.41$ kJ mol$^{-1}$
• B. $-0.21$ kJ mol$^{-1}$
• C. $+0.41$ kJ mol$^{-1}$
• D. $-0.14$ kJ mol$^{-1}$
• E. $+0.32$ kJ mol$^{-1}$

Hint:

Always convert entropy from J to kJ when enthalpy is given in kJ.

Answer 1

The Correct Option is C

Concept: Chemistry - Gibbs Free Energy. [ \Delta G=\Delta H-T\Delta S ]
Step 1: Convert entropy unit. [ \Delta S=-45\text{ J K}^{-1}\text{mol}^{-1}=-0.045\text{ kJ K}^{-1}\text{mol}^{-1} ]
Step 2: Substitute values. [ \Delta G=-13-298(-0.045) ]
Step 3: Calculate. [ \Delta G=-13+13.41=0.41 ] [ \Delta G=+0.41\text{ kJ mol}^{-1} ]
Step 4: Final answer. [ \boxed{+0.41\text{ kJ mol}^{-1}} ]
Hence, correct option is (C).