Define a dynamic equilibrium - NSC Physical Sciences - Question 6 - 2018 - Paper 2

From the graph, at t = 2 minutes, the value of x is 8 moles.

At the same time, the value of y is obtained from the graph, which shows 4 moles.

The value of z from the graph at equilibrium is 12 moles.

To calculate the equilibrium constant Kc, use the formula:

$K_c = \frac{[C]}{[A]^a[B]^b}$

Given:

• Volume = 3 dm³
• At equilibrium: z = 12 moles, x = 8 moles, y = 4 moles

Concentrations at equilibrium:

• $[C] = \frac{z}{3} = \frac{12}{3} = 4 \text{ mol/dm}^3$
• $[A] = \frac{x}{3} = \frac{8}{3} = 2.67 \text{ mol/dm}^3$
• $[B] = \frac{y}{3} = \frac{4}{3} = 1.33 \text{ mol/dm}^3$

Substituting the values into the equation:

$K_c = \frac{4}{(2.67)^{3}(1.33)} = 6.75$

The reverse reaction is ENDOTHERMIC. An increase in temperature typically favors the endothermic reaction, as it absorbs heat from the surroundings. Therefore, as the temperature increases, it shifts the equilibrium position towards the side of the endothermic reaction.