4. assuming the reaction below is at equilibrium, which of the following changes will drive the reaction to the left? c(s) + o2(g) + co2(g) ah -393.5 kj/mol 1. increasing the temperature 2. adding o2(g) 3. removing c(s) a. i only b. 2 only c. 3 only d. 1 and 3 e. 2 and 3

The answer is 1. Increasing the temperature.

Explanation:

When changes are applied to a system that is at equilibrium it will respond according to Le Chatelier´s Principle. This principle states that changes in temperature, volume, pressure or concentration will cause the system to react in a way that opposes the change, trying to achieve a new equilibrium state.

In 1. we need to take into account that this is an exothermic reaction (ΔH° is negative). Therefore, we could imagine heat as one of the products.

C(s) + O₂(g) ⇄ CO₂(g) + heat

If we increase the temperature, the system will react trying to reduce it. To do so, it has to consume heat, that is, shift towards the reactants (left).

In 2, if we add O₂(g), its concentration increases and the system will try to decrease it by shifting the reaction towards the products (right) and consuming the O₂(g).

In 3, solids do not form part of the equilibrium (they are not in the equilibrium constant) because their concentration doesn´t change much over time. Therefore, adding or removing solid C(s) makes no change.

All in all, only 1. (increasing the temperature) will drive the reaction to the left.

50.23 i believe is correct but don't quote me on that.