Carbon, C(s), reacts with sulphur, S(g), according to the following balanced equation: C(s) + 2S(g) ⇌ CS2(g) ΔH > 0 The system reaches equilibrium at temperature T in a sealed 2 dm³ container - NSC Physical Sciences - Question 6 - 2022 - Paper 2

Given the balanced equation: C(s) + 2S(g) ⇌ CS2(g)

At equilibrium, we have 1 mole of CS2. Based on stoichiometry, 2 moles of S(g) are needed for every mole of CS2 formed. Therefore, the initial amount of S(g) is:

• Initial moles of S(g) = 2 mol for 1 mol of CS2 at equilibrium.

Calculating the concentration of S(g):

Concentration (C) of a gas is given by the formula: $C = \frac{n}{V}$

Where:

• n = amount of substance in moles
• V = volume in dm³

For S(g), we have:

• n = 2 moles of S
• V = 2 dm³ $C_{S} = \frac{2}{2} = 1 ext{ mol·dm}^{-3}$

The amount of S(g) will DECREASE as the new equilibrium is established. This is because the volume is doubled, leading to a decrease in pressure which shifts the equilibrium position to favor the side with more moles of gas, hence consuming S(g).

According to Le Chatelier's principle, if the volume of the container is increased, the system will try to counter this change by favoring the reaction that produces more gas molecules. In this case, the forward reaction is favored as the system works to establish a new equilibrium, resulting in a decrease in the amount of S(g).

The equilibrium expression for the reaction is given by: $K_c = \frac{[CS_2]}{[S]^2}$

If the concentration of CS2(g) changes by x mol·dm⁻³, the new concentration of CS2(g) will be: $[CS_2] = 1 + x$

Since the concentration of S(g) will decrease due to the shifting equilibrium: $[S] = 1 - 0.5x$

Thus, we can express the equilibrium constant as: $K_c = \frac{1+x}{(1-0.5x)^2}$

The parallel lines between tA and tB represent a state of equilibrium where the rates of the forward and reverse reactions are equal. During this time, there is no net change in the concentration of the reactants or products.

At tB, an external change was made to the system, such as an increase in temperature or concentration of reactants/products, leading to a shift in the equilibrium position.

The sudden change in reaction rate at tC can be attributed to a change in conditions, such as a shift in equilibrium due to an increase in temperature or a change in concentration of reactants/products, affecting the rates of the forward and reverse reactions.

As stated in QUESTION 6.3, the amount of S(g) decreases because the system responds to the change in volume. By Le Chatelier's principle, increasing the volume decreases the pressure, prompting the reaction to favor the side with more gas molecules (in this instance, it moves towards the reactants). Hence, the equilibrium adjusts to favor the production of more S(g), leading to its overall decrease in the system.