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ΔG∘=−RTln⁡K\Delta G^{\circ} = -RT\ln K

Why this formula appears here

The reason is a bookkeeping identity rather than an empirical observation. The equilibrium constant is fixed by the standard Gibbs energy difference between reactants and products, ΔG∘=−RTln⁡K\Delta G^{\circ} = -RT\ln K. and that difference is a property of the two end states alone. A catalyst is present in the initial state and present again, unaltered, in the final state. It therefore cancels out of the difference. Any device that genuinely shifted an equilibrium while returning to its original condition would be a perpetual motion machine of the second kind, and could be operated in a loop to extract work from a single heat bath.

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ΔG∘\Delta G^{\circ}

Symbol Δ G^circ

Δ GcG^circ is part of the quantity the equation computes from the expression on the right.

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Published contexts (1)

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ΔG∘=−RTln⁡K,\Delta G^{\circ} = -RT\ln K,

Equation 1 · Chemistry & Catalysis

Lowering the Barrier: What a Catalyst Actually Does

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions.

The reason is a bookkeeping identity rather than an empirical observation. The equilibrium constant is fixed by the standard Gibbs energy difference between reactants and products, ΔG∘=−RTln⁡K\Delta G^{\circ} = -RT\ln K. and that difference is a property of the two end states alone. A catalyst is present in the initial state and present again, unaltered, in the final state. It therefore cancels out of the difference. Any device that genuinely shifted an equilibrium while returning to its original condition would be a perpetual motion machine of the second kind, and could be operated in a loop to extract work from a single heat bath.

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