Consider the following system, which is at equilibrium, 3C(s) + 3H2(g) <--> CH4(g) + C2H2(g) The result of removing some C(s) from the system will be:

The equilibrium position will shift to the left, thus, favouring the backward reaction.

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The equilibrium position will shift to the left, thus, favouring the backward reaction.

Explanation:

The equation for the reaction is given below:

3C(s) + 3H2(g) <==> CH4(g) + C2H2(g)

According to Le Chatelier's principle, if an external constrain such as change in concentration, temperature or pressure is imposed on a chemical system in equilibrium, the equilibrium will shift in order to neutralize the effect.

From Le Chatelier's principle, removing some C(s) implies removing some of the concentration of the reactants.

This will shift the equilibrium position to the left, thus, favouring the backward reaction (i.e forming more reactants) because removing C(s) implies that the products are now more than the reactants and as such, they will react to form more reactants.

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because it does not have the potential to be entirely used up

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.