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According to the equation 33 above, shouldn't be all the reactions be spontaneous initially? Because the reaction quotient Q, is zero at the start of the reaction and logarithm of Q makes it a very large negative value and ultimately we'll have ∆G to be a large, negative value which makes the reaction spontaneous?

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Your interpretation of the equation is correct. $\Delta G^o$ gives you thermodynamic favorability of a reaction under standard conditions (Q=1) and even reactions with positive values of $\Delta G^o$ (unfavored under standard conditions) can be driven to proceed if the concentrations of the reactants and products are made extreme enough.

The best way to think about the process is via a curve of G vs reaction extent $\xi$, this is related to Q and ranges from 0 (pure reactants, Q = 0, #1 on the graph) to 1 (pure products, Q = ∞, # 2 on the graph):

enter image description here

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/23.4%3A_Free_Energy_and_the_Gibbs_Function

Any reaction will have an equilibrium point on this curve, corresponding to a most stable composition (# 3 on the graph). The value of the equilibrium constant (K) tells us the x coordinate of the most stable position (i.e. how much reactant vs product we have in the equilibrium mixture).

No matter what the value of K, it will always be in the range $0<K<∞$. If we start the process with pure reactants, $Q = 0$ and we are to the left of the equilibrium point. The system will shift to the right and some of the reactants will convert into products.

This is what you mean by

ultimately we'll have ∆G to be a large, negative value which makes the reaction spontaneous?

you can calculate a negative $\Delta G{_r}$ for the reaction when you have no (or infinitesimal amounts of) products.

Many texts are moving away from the term "spontaneous" for this situation, because the system may not shift to the right (reactants convert to products) if there is a kinetic barrier to prevent this, so the forward reaction may not actually be observed. The term "thermodynamically favored" is often used instead, meaning having a negative $\Delta G{_r}$ under these conditions.

You can find lots more about these concepts at the following posts:

If change in free energy (G) is positive, how do those reactions still occur?

Calculating ΔG at the extremes of reaction extent

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What you write is correct about all reactions proceeding if starting with reactants only, these will decrease in concentration and products increase until equilibrium is reached. How long this takes is a matter for chemical kinetics not thermodynamics. Strictly speaking thermodynamics has nothing to say about this process as it deals only with equilibrium situations. Think of the word 'spontaneous' in thermodynamics as not having the same meaning as the word used in general language, but describes only that $K_p > 1$ so that $\Delta G^{\mathrm{0}}<0 $.

The graph below shows $\Delta G=\Delta G^\mathrm{0} +RT\ln(Q)$ where for a reaction $\ce{A <=>B}$ at equilibrium $\displaystyle Q\equiv K_p=\frac{p_B}{p_A}$ where $p$'s are the partial pressures. The $\Delta G$ is the gradient of $G$ with extent of reaction $\xi$ which is zero when only reactants are present and is $1$ when $1$ mole of reactants have been converted to products. Thus $\Delta G$ is the slope of the curve shown in the plot in the previous answer.

An alternative way of looking at this is shown in the figure below. You can see that as $Q$ becomes smaller $\Delta G$ becomes larger and negative, and vice versa. The slope of the plot is $RT$.

deltag

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