Chemistry - Transition state and free energy
Solution 1:
According to the principle of Microscopic Reversibility all elementary reactions, i.e. those that proceed in a single step, are reversible. And you are right: For a reversible reaction $\Delta G$ is very small while for an irreversible reaction $\Delta G$ is large. This can be most easily described using a simple schematic energy curve diagram for your situation.
Each reaction is microscopically reversible. So either your reactants $\ce{A}$ and $\ce{B}$ or your products $\ce{C}$ and $\ce{D}$ can combine forming the transition state ($\ce{TS}$). As your text book says, the $\ce{TS}$, being an energy maximum, is very unstable and collapses very fast. Since the transition state represents a maximum of the energy curve the collapse may go in either the direction of the reactants or the direction of the products with equal probability, pretty much in the same way a ball placed at the top of hill may roll down on either side of it. So, once you got to the transition state you have a 50:50 situation. The thing that determines which way your reaction goes is how easy it is to make the transition state or to "get the ball to the top". In order reach the transition state you have to overcome a thermodynamic barrier, which is the difference of the free energies of products and reactants $\Delta G$, and a kinetic barrier, which is the activation energy $E_{\mathrm{a}}$. If $\Delta G$ is a very large and negative then the reaction will proceed in the direction from $\ce{A}+\ce{B}$ to $\ce{C}+\ce{D}$ and will be irreversible because then it will take considerably more effort to "climb" the energy hill coming from the $\ce{C}+\ce{D}$ side than coming from the $\ce{A}+\ce{B}$ side. This means that in a given time a lot more $\ce{A}$ and $\ce{B}$ molecules are "elevated" to the $\ce{TS}$ than $\ce{C}$ and $\ce{D}$ molecules. So, even if $\ce{TS}$ will revert to $\ce{A} + \ce{B}$ 50 % of the time the reverted molecules will quickly be back "on top of the hill" in order to have another try while for $\ce{C}$ and $\ce{D}$ this takes much longer - the reaction rate for the forward reaction is much higher than the reaction rate for the backward reaction. So over time more and more $\ce{A} + \ce{B}$ is converted to $\ce{C}+\ce{D}$ until equilibrium is estabilished and the ratio of reactants/products is equal to the ratio of the reaction rates. When you have a reversible reaction and $\Delta G$ is very small the "hill to climb to the transition state" is nearly the same height irrespective of whether you come from the product or from the reactant side. So, forward and backward reaction proceed nearly at the same rate.
Concerning your second question: I'm sorry, but I couldn't find a non-mathematical explanation as to why $G$ depends on the concentration. The Gibbs free energy is the thermodynamic potential of an ensemble at constant pressure and temperature. For variable particle numbers $N_{i}$ its fundamental relation is:
\begin{equation} dG = -S dT + pdV + \sum_{i} \mu_{i} dN_{i} \ . \end{equation}
So, it is connected to the particle numbers (and thus to the concentrations) of the species in the system.
Solution 2:
First you are correct, there is no fundamental difference in reactions being described as reversible or irreversible, unlike in thermodynamics. A reaction will be called irreversible (a)if the product is removed from reaction i.e. by precipitation or physical removal and (b) if the rate of returning from product to reactants is so slow that it cannot be measured. This is somewhat subjective, but the lifetime of the back reaction can be years, so in practice its clear cut.
Secondly, the equilibrium constant for a reaction $\ce{aA + bB <=>cC + dD} $ is defined as $$K_p=\frac{P_C^cP_D^d}{P_A^aP_B^b}$$ where $P_A^b$ etc. represents the pressure at equilibrium. As the equilibrium constant must be dimensionless so pressure is understood to be divided by unit pressure of 1 bar. If the reaction is in solution then concentrations are used instead of pressure, then these are understood to be divided by $\ce{1 mol~dm^{-3}}$ (Technically instead of pressure or concentrations activities are used but for most uses pressure and concentration are sufficient). The standard free energy is defined as$$ \Delta G^0 = -RTln(K_p)$$ and this is how the pressures/concentrations come into the equilibrium constant.
The transition state is reached only when the reactants have sufficient energy to do so. By the Boltzmann distribution the number of energetic molecules rapidly diminishes as the energy is increased. Thus most collisions do not result in reaction, unless the activation barrier is very small compared to average (thermal ) energy. The reverse reaction has a larger barrier to surmount if the reaction is exothermic, thus fewer collisions result in reaction back to products than the other way round.
The transition state is at the top of the barrier between reactants and products, in solution it may be crossed and recrossed many times before reaction is complete, it all depends upon how quickly the energy the 'molecule' has at the transition state can be removed by collisions or by energy being redistributed internally into the products as vibrations and rotations or as kinetic energy of separation.
The activation energy barrier exists because the product will have different arrangement of bonds and atoms from the reactant and so different average internuclear separation in the reaction coordinate. The activation energy can be imagined as the point where two potential energy parabolas cross, one being slightly displaced in position and energy vs. the other. This is what the normal reaction profile drawn in text books and elsewhere represents. It is usually shown as one line(the bottom of the first parabola), a curved line to a maximum (crossing point, transition state) then onto another lower line to represent product energy at the minima of the second parabola.
Solution 3:
First question: The answer is true from a reaction view. You are right. No matter how the reaction is favored in terms of Gibbs free energy change. The reverse process will happen in small or tiny amount. From a thermodynamics view, they are different. Similar to the reversible and irreversible process in physics, reversible reaction has no energy loss but irreversible reaction has energy loss.
Second question: Gibbs free energy change is a function depending on concentration for solution and partial pressure for gas. So concentration will obviously affect it. Do not confuse it with the standard Gibbs free energy change, which assume that you have a standard condition (1 Pa pressure or 1 mol/L concentration). Since the concentration or partial pressure is a constant (precondition), the standard Gibbs free energy change is independent to concentration or partial pressure.
Both of your question is not related to transition state. Transition state is part of the chemical kinetics, not thermodynamics.