<P> Q is the reaction quotient . At equilibrium, Δ G = 0, and Q = K, so the equation becomes Δ G _̊ = − RT ln K; K is the equilibrium constant . </P> <Table> Table of selected substances <Tr> <Th> Substance <P> (State) </P> </Th> <Th> Δ G ° <P> (kJ / mol) </P> </Th> <Th> Δ G ° <P> (kcal / mol) </P> </Th> </Tr> <Tr> <Td> NO <P> (g) </P> </Td> <Td> 87.6 </Td> <Td> 20.9 </Td> </Tr> <Tr> <Td> NO <P> (g) </P> </Td> <Td> 51.3 </Td> <Td> 12.3 </Td> </Tr> <Tr> <Td> N O <P> (g) </P> </Td> <Td> 103.7 </Td> <Td> 24.78 </Td> </Tr> <Tr> <Td> H O <P> (g) </P> </Td> <Td> − 228.6 </Td> <Td> − 54.64 </Td> </Tr> <Tr> <Td> H O <P> (l) </P> </Td> <Td> − 237.1 </Td> <Td> − 56.67 </Td> </Tr> <Tr> <Td> CO <P> (g) </P> </Td> <Td> − 394.4 </Td> <Td> − 94.26 </Td> </Tr> <Tr> <Td> CO <P> (g) </P> </Td> <Td> − 137.2 </Td> <Td> − 32.79 </Td> </Tr> <Tr> <Td> CH <P> (g) </P> </Td> <Td> − 50.5 </Td> <Td> − 12.1 </Td> </Tr> <Tr> <Td> <P> (g) </P> </Td> <Td> − 32.0 </Td> <Td> − 7.65 </Td> </Tr> <Tr> <Td> <P> (g) </P> </Td> <Td> − 23.4 </Td> <Td> − 5.59 </Td> </Tr> <Tr> <Td> <P> (g) </P> </Td> <Td> 129.7 </Td> <Td> 29.76 </Td> </Tr> <Tr> <Td> <P> (l) </P> </Td> <Td> 124.5 </Td> <Td> 31.00 </Td> </Tr> </Table> <Tr> <Th> Substance <P> (State) </P> </Th> <Th> Δ G ° <P> (kJ / mol) </P> </Th> <Th> Δ G ° <P> (kcal / mol) </P> </Th> </Tr> <P> (State) </P>

What information does gibbs free energy give about a reaction answers