Correct Answer: Option A

Explanation

For a reaction to be feasible (spontaneous), the change in Gibbs free energy (ΔG) must be negative (ΔG<0). The Gibbs free energy is related to the enthalpy change (ΔH) and entropy change (ΔS) by the equation:  ΔG = ΔH −TΔS

In the given reaction, 2SO\(_2\)\(_{(s)}\)  +  O\(_2\)\(_{(s)}\) ⇌  2SO\(_3\), there are 3 moles of solid reactants forming 2 moles of solid products. A decrease in the number of moles of solid generally leads to a decrease in entropy, so ΔS would be negative. The formation of SO\(_3\) from SO\(_2\) and O\(_2\) is an exothermic reaction, meaning ΔH is negative. 

When ΔH is negative and ΔS is negative, the reaction is feasible at low temperatures. 

Based on the options provided: 

A. ΔH = +ve, ΔS = +ve; Feasible only at high temperatures.

B. Δ? = 0, ΔS= + ve ; Feasible at all temperatures as ΔG would be negative.

C. ΔH = +ve , ΔS = −ve : This combination always results in a positive ΔG, so the reaction is never feasible in the forward direction as written.

D. ΔS = 0: Feasible if ΔH is negative. The condition which makes any reaction not feasible is a positive ΔH and a negative ΔS. 

If both are positive, low T makes −TΔS negative, so high T favours negative ΔG (spontaneous). 

There is an explanation video available below.

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