For a reaction to be feasible, the value of ∆G must be zero or negative.

The idea

The test for feasibility is the sign of ∆G: a reaction is feasible when ∆G is zero or negative, and is not feasible when ∆G is positive. Applying it to ammonium nitrate dissolving at 298 K, with ∆H of +26 kJ mol⁻¹ and ∆S of +108 J K⁻¹ mol⁻¹, gives ∆G = 26 − (298 × 0.108) = −6.2 kJ mol⁻¹, so the process is feasible despite being endothermic. Feasible is a thermodynamic statement only. It says the reaction is energetically allowed, not that it will be observed: a reaction with a large activation energy can have a strongly negative ∆G and still proceed imperceptibly slowly, which is why diamond does not turn into graphite while you watch it.

What earns the marks

• Stating that a reaction is feasible when ∆G is zero or negative.
• Calculating ∆G and quoting the sign explicitly before drawing a conclusion.
• Concluding feasible or not feasible consistently with the calculated value, allowing error carried forward.
• Recognising that a feasible reaction may still be too slow to observe because of a high activation energy.

Common mistakes and how to put them right

• Concluding that a reaction is feasible whenever ∆H is negative, without calculating ∆G.

• Reading a positive ∆G as feasible because the number looks small.

• Calculating ∆G correctly and then drawing the opposite conclusion from its sign.

• Claiming a reaction with negative ∆G must be fast.

• Stating that ∆G must be negative, overlooking that zero is the point at which it becomes feasible.

Exam tips

• Always write one sentence linking your calculated value to feasibility; the conclusion is a separate mark.
• If your ∆G is positive, say not feasible at this temperature rather than not feasible at all.
• If a question asks why a feasible reaction is not observed, answer with a high activation energy, not with entropy.