Question

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

Answer 1

Increasing temperature

Step-by-step explanation:

`C(s) + O_2(g) \leftrightharpoons CO_2(g)`

Enthalpy of the reaction = -393.5 kJ/mol

Negative sign implies that reaction is exothermic.

Effect of change in reaction condition is explained by Le chateliers principle.

According to Le chateliers principle, if the reaction conditions of a reversible reaction in a state of dynamic equilibirum is changed, the reaction will move in a direction to counteract the change.

1. Increasing the temperature

Forward reaction is exothermic that means temperature increases in forward direction. Backward reaction will be endothermic and so there is decrease in temperature in backward direction or in left direction.

On increasing temperature, reaction will be move in direction to counteract the increased temperature, therefore reaction will move in left direction.

2. Adding O2

If O2 is added, then reaction will move in a direction in which its get consumed. So, reaction will move in forward direction or in right direction.

3. Removing C (s)

Le Chatelier's principle does not apply on solids, so removal of C(s) does not affect the equilibrium.

Answer 2

The answer is 1. Increasing the temperature.

Step-by-step 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.