2SO\(_2\)\(_{(s)}\) + O\(_2\)\(_{(s)}\) ⇌ 2SO\(_3\) ; ΔG° = - ve For the above reaction to be feasible

A reaction is feasible (spontaneous) when the Gibbs free energy change is negative: \(\Delta G < 0\). The relationship between Gibbs free energy, enthalpy, and entropy is:

\[\Delta G = \Delta H - T\Delta S\]

The question states that \(\Delta G^\circ\) is negative. To determine which combination of \(\Delta H\) and \(\Delta S\) guarantees this, consider each option:

• If \(\Delta H\) is positive and \(\Delta S\) is positive: \(\Delta G = (+) - T(+)\). This is negative only at sufficiently high temperatures, so feasibility is not guaranteed at all temperatures.
• If \(\Delta H = 0\) and \(\Delta S\) is positive: \(\Delta G = 0 - T(+) = -T\Delta S\), which is always negative at any temperature above 0 K. This guarantees a negative \(\Delta G\).
• If \(\Delta H\) is positive and \(\Delta S\) is negative: \(\Delta G = (+) - T(-) = (+) + T|\Delta S|\), which is always positive. The reaction would never be feasible.
• If \(\Delta S = 0\): \(\Delta G = \Delta H\), and feasibility depends entirely on the sign of \(\Delta H\), which is not specified.

The only option that ensures \(\Delta G\) is negative under all conditions is \(\Delta H = 0\) and \(\Delta S\) is positive, because the \(-T\Delta S\) term is always negative when \(\Delta S > 0\).