Question

When sulphur (in the form of \(S_{8}\) ) is heated at temperature \(T\), at equilibrium, the pressure of \(S_{8}\) falls by \(30 \%\) from \(1.0 \mathrm{~atm}\), because \(\mathrm{S}_{8}(g)\) is partially converted into \(\mathrm{S}_{2}(g)\). Find the value of \(K_{p}\) for this reaction. (a) \(2.96\) (b) \(6.14\) (c) \(204.8\) (d) None of these

Short answer

2.96

Step-by-step solution

- Understanding the chemical reaction

Assume initially there's 1 mole of \(S_8\) gas. During the process at equilibrium, as the pressure of \(S_8\) falls by 30\%, it means that 30\% of \(S_8\) has been converted into \(S_2\). The chemical equilibrium reaction can be expressed as: \(\text{S}_8(g) \rightleftharpoons 4\text{S}_2(g)\).

- Calculating the change in moles at equilibrium

If initially there is 1 mole of \(S_8\), at equilibrium there will be \((1 - 0.3) = 0.7\) moles of \(S_8\) left. Since for every mole of \(S_8\) that reacts, 4 moles of \(S_2\) are produced, the moles of \(S_2\) formed at equilibrium will be \((4 \times 0.3) = 1.2\) moles.

- Writing the expression for \(K_p\)

The expression for the equilibrium constant in terms of partial pressures \(K_p\) is given by \(K_p = \frac{(P_{S_{2}})^4}{P_{S_{8}}}\). Since the partial pressures are proportional to moles for a given volume and temperature, we can substitute moles for partial pressures in the \(K_p\) expression.

- Find the initial partial pressures

The initial partial pressure of \(S_8\) is 1 atm. Since no \(S_2\) is present initially, its partial pressure is 0 atm.

- Calculating the partial pressures at equilibrium

At equilibrium, the partial pressure of \(S_8\) is 30\% less than the initial value, so it's \((1.0 \, \text{atm} - 0.3 \times 1.0 \text{atm}) = 0.7 \, \text{atm}\). As the volume and temperature remain constant, the change in moles of \(S_2\) is directly proportional to the change in its partial pressure. Thus the partial pressure of \(S_2\) is \((1.2/1.0) \times 1.0 \, \text{atm} = 1.2 \, \text{atm}\).

- Plug in the values into the \(K_p\) expression

Insert the calculated equilibrium partial pressures in the \(K_p\) expression: \(K_p = \frac{(1.2 \, \text{atm})^4}{0.7 \, \text{atm}}\).

- Calculate \(K_p\)

Perform the calculations: \(K_p = \frac{1.2^4 \, \text{atm}^4}{0.7 \, \text{atm}} = \frac{2.0736 \, \text{atm}^4}{0.7 \, \text{atm}} = 2.96229 \, \text{atm}^3\). This value rounds off to 2.96.

Key concepts

Kp Calculation

Understanding how to calculate the equilibrium constant, represented as K_p, is crucial when exploring chemical equilibrium. K_p is based on the partial pressures of gases at equilibrium. For a balanced chemical equation of the form aA(g) + bB(g) ⇌ cC(g) + dD(g), K_p is determined using the expression:
K_p = \(\frac{(P_C)^c \cdot (P_D)^d}{(P_A)^a \cdot (P_B)^b}\) where P represents the partial pressures of components in atmosphere (atm) and a, b, c, d denote the stoichiometric coefficients. To solve for K_p, one must know or calculate the partial pressures of the reactants and products at the equilibrium state. For gaseous equilibria, as in the exercise, this involves a direct relation between the drop in the reactant's pressure and the formation of products' pressure.

Partial Pressure in Equilibrium

Partial pressure describes the pressure that each gas in a mixture would exert if it were alone in a container. At equilibrium, the sum of the partial pressures of all gases equals the total pressure. In our exercise, the initial total pressure is exerted solely by S₈(g), and as it partially converts to S₂(g), the partial pressures of both gases at equilibrium need to be accounted for. A 30% decline in the S₈ pressure after equilibrium suggests a corresponding increase in the S₂ pressure. The proportionality becomes handy as we can convert moles to pressure directly when volume and temperature are constant. This way, the partial pressures reflect the mole fractions of the gases, allowing one to apply them in the K_p expression to find the equilibrium constant.

Le Chatelier's Principle

Le Chatelier's principle helps predict how a system at equilibrium reacts to changes in concentration, temperature, or pressure. It states that if a dynamic equilibrium is disturbed, the system adjusts in a way that counteracts the change. For example, if the pressure were increased in the given exercise, the system would shift to reduce pressure, typically by favoring the reaction that produces fewer moles of gas. Consequently, understanding the shifts can provide insights on how the equilibrium concentrations or pressures might change. This adjustment is key to predicting how K_p may vary with the conditions and helps students understand that equilibrium states are not static but dynamic and responsive to external influences.

Mole-to-Mole Conversion in Chemical Reactions

In chemical reactions, the stoichiometry plays an integral role in establishing relationships between reactants and products. Mole-to-mole conversion is a process of using the balanced equation to understand these relationships. For instance, the reaction provided shows that 1 mole of S₈(g) produces 4 moles of S₂(g). Therefore, the conversion of S₈ moles at equilibrium to S₂ is pivotal for calculating the partial pressures that are used to find K_p. Understanding these conversions guides in determining how much of a reactant is needed to form a desired amount of product and is essential in quantifying the components at equilibrium.