Let's analyze the given reaction and the effects of the changes on the equilibrium:
[tex]\[ N_2(g) + 2O_2(g) \rightleftharpoons 2NO_2(g) \][/tex]
Given that the forward reaction is endothermic, it absorbs heat. Hence, increasing the temperature shifts the equilibrium towards the products to absorb the extra heat.
Next, let's consider the effect of pressure. According to Le Chatelier's principle, increasing the pressure of a reaction involving gases will shift the equilibrium to the side with fewer moles of gas. In this reaction, there are 3 moles of gas on the reactant side (1 mole of N_2 and 2 moles of O_2) and 2 moles on the product side (2 moles of NO_2). Therefore, increasing the pressure will shift the equilibrium towards the products.
Now, let's examine each of the given changes:
I. Increase NO_2 - Increasing the concentration of a product will shift the equilibrium towards the reactants to relieve the added stress (Le Chatelier's principle).
II. Decrease O_2 - Decreasing the concentration of a reactant will shift the equilibrium towards the reactants to counteract the reduction.
III. Add a catalyst - A catalyst speeds up the attainment of equilibrium but does not change the position of the equilibrium.
IV. Increase the temperature - For an endothermic reaction, increasing the temperature shifts the equilibrium towards the products.
V. Increase the pressure - Since the product side has fewer moles of gas, increasing the pressure will shift the equilibrium towards the products.
Summarizing the effects:
- IV (Increase the temperature) shifts the equilibrium towards the products.
- V (Increase the pressure) shifts the equilibrium towards the products.
Thus, the correct answer is:
A. IV and V
