Consider the following reaction at equilibrium. what effect will adding some c have on the system? co2(g) + c(graphite) ↔ 2 co(g)the equilibrium constant will increase. no effect will be observed since c is not included in the equilibrium expression. the equilibrium constant will decrease. the reaction will shift to the left in the direction of reactants. the reaction will shift to the right in the direction of products.

No effect will be observed since C is not included in the equilibrium expression.

Explanation:

Let's consider the following reaction.

CO₂(g) + C(graphite) ↔ 2 CO(g)

The equilibrium constant (Kc) is the product of the concentration of the products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants raised to their stoichiometric coefficients.

Kc only includes gases and aqueous species (not liquids or solids).

As we can see, C is a solid and is not included in the equilibrium expression, so adding some C would have no effect on it.

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.