Consider the following reversible reaction $$2NO(g) \rightleftharpoons N_2(g) + O_2(g) \quad \Delta H = -183 \text{ kJ}$$ that has an equilibrium constant (Kc) value of 20.25 at a certain high temperature T: (a) Write the equilibrium constant expression for the reaction - Leaving Cert Chemistry - Question 9 - 2023

Given that the equilibrium constant Kc is 20.25, we set up an ICE table:

The equilibrium expression gives us:

$K_c = \frac{[N_2][O_2]}{[NO]^2} = \frac{x \cdot x}{(2 - 2x)^2} = 20.25$

Substituting in for Kc:

$20.25 = \frac{x^2}{(2 - 2x)^2}$

Cross-multiplying and solving:

$20.25(2 - 2x)^2 = x^2$

Letting ( x = 0.9 \text{ moles} ), we find: The number of moles of nitrogen gas (N2) at equilibrium is thus 0.9 moles.

Le Chatelier's principle states that if a system at equilibrium is subjected to a change (stress), the system will adjust to counteract that change and restore a new equilibrium.

An increase in temperature for this exothermic reaction will result in the equilibrium shifting to the left (toward the reactants) to absorb the added heat. Therefore, the value of Kc decreases.

An increase in pressure will shift the equilibrium toward the side with fewer moles of gas. In this case, the left side has 2 moles (2NO) while the right side has 2 moles (1 N2 + 1 O2). There is no change on the side of the equilibrium due to pressure, meaning Kc will remain the same.