Question

what dot it mean when delta G=-ve and delta G =+be?​

Answer 1

Answer:We define ΔG0' (pronounced “delta G naught prime”) as the free energy change of a reaction under “standard conditions” which are defined as: All reactants and products are at an initial concentration of 1.0M. Pressure of 1.0 atm.

Explanation:∆G is the change of Gibbs (free) energy for a system and ∆G° is the Gibbs energy change for a system under standard conditions (1 atm, 298K). ... Where ∆G is the difference in the energy between reactants and products. In addition ∆G is unaffected by external factors that change the kinetics of the reaction.

Answer 2

Answer:

• Standard condition means the pressure 1 bar and Temp 298K, ΔG° is the measure of Gibbs Free Energy (G) - The energy associated with a chemical reaction that can be used to do work change at 1 bar and 298 K, delta G "naught" (not) is NOT necessarily a non-zero value. ΔG° = -RT ln(K), So ΔG° = 0, if K = 1. In other words, if the product and reactant are equally favored at equilibrium, it's because there is no difference speaking. ΔG° is always the same for a given reaction. ΔG does depend on your conditions but still relates to ΔH° and ΔS°. Consider the following:
• ΔG = ΔG° + RT ln(P) Where P is the reaction quotient, the ratio of products to reactants at some state. It is equal to K if the system has reached equilibrium. Subbing in ΔG°=ΔH°-TΔS°,
• ΔG = ΔH°-TΔS° + RT ln(P)

• So, ΔG does relate to those two quantities. ΔH° and ΔS° represent the change in enthalpy and entropy between product and reactant. but they DO NOT mean a "100% complete reaction." They determine the energetic difference (at a given temperature). This difference in energy determines the composition at equilibrium. It has nothing to do with the "completeness" of the reaction, which is a kinetic question. So now in most questions, such as in phase changes questions, they would give us the ΔH° and ΔS° and they then would want us to find the temperature at which equilibrium takes place. So using the formula ΔG°=ΔH°-TΔS° we would let ΔG° be 0 and solve for T. ΔG° is a non-zero value and that we shouldn't be able to use ΔH° or ΔS° to find ΔG because either ΔH° or ΔS° represents 100% complete reaction.