Question

Consider the following equilibrium: \(2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g) \quad K_{c}=1.08 \times 10^{7}\) at \(700^{\circ} \mathrm{C}\) (a) Calculate \(K_{p}\). (b) Does the equilibrium mixture contain mostly \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) or mostly \(\mathrm{H}_{2} \mathrm{~S} ?\) (c) Calculate the values of \(\mathrm{K}_{c}\) and \(K_{p}\) if you rewrote the balanced chemical equation with \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2}(g)\) instead of \(2 \mathrm{~mol}\).

Short answer

In summary, for the given equilibrium reaction, we find: a) The value of \(K_p\) is approximately \(5.02 \times 10^{5}\). b) The equilibrium mixture contains mostly \(\mathrm{H}_{2}\mathrm{S}\), due to the large value of \(K_c\). c) For the modified balanced chemical equation, the new values of \(K'_c\) and \(K'_p\) are approximately \(3.28 \times 10^3\) and \(3.55 \times 10^2\), respectively.

Step-by-step solution

Recall the relationship between K_c and K_p

We can calculate \(K_p\) using the following equation: \(K_p = K_c(RT)^{\Delta{ n}}\) where: - \(K_p\) is the equilibrium constant in terms of pressure - \(K_c\) is the equilibrium constant in terms of concentration, which is given as \(1.08 \times 10^{7}\) - \(R\) is the ideal gas constant, which is \(8.314~\mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\) - \(T\) is the temperature in Kelvin - \(\Delta{n}\) is the change in the number of moles of gas in the reaction

Find the change in the number of moles of gas, Δn

Calculate the change in the number of moles of gas in the reaction: \(\Delta n = (\text{moles of product}) - (\text{moles of reactant})\) For the given reaction: \(\Delta n = (2~\text{moles of}\, \mathrm{H}_{2} \mathrm{~S}(g)) - (2~\text{moles of}\, \mathrm{H}_{2}(g) + 1~\text{mole of}\, \mathrm{S}_{2}(g))\) \(\Delta n = 2 - 3 = -1\)

Convert the temperature to Kelvin

We need to convert the given temperature, which is \(700^{\circ}\mathrm{C}\), to Kelvin: \(T(K) = 700 + 273.15\) \(T(K) = 973.15~\mathrm{K}\)

Calculate K_p

Replace the constants and variables values in the equation: \(K_p = (1.08 \times 10^{7})(8.314)(973.15)^{-1}\) Now, solve for \(K_p\): \(K_p \approx 5.02 \times 10^{5}\) So, the value of \(K_p\) is approximately \(5.02 \times 10^{5}\). #b) Composition of the Equilibrium Mixture#

Analyze the value of K_c

The given value of \(K_c\) is \(1.08 \times 10^7\), which is a large number. A large value for \(K_c\) indicates that at equilibrium, the concentration of products is higher than the concentration of reactants. In this case, the equilibrium mixture contains mostly \(\mathrm{H}_{2}\mathrm{S}\). #c) Calculating new K_c and K_p values for the modified balanced chemical equation#

Write the modified balanced chemical equation

The modified balanced chemical equation contains \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2}(g)\) instead of \(2 \mathrm{~mol}\): \(\mathrm{H}_{2}(g)+\frac{1}{2}\mathrm{S}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)\)

Calculate the new K_c value

The new balanced chemical equation is 1/2 the original equation. Thus, the new \(K'_c\) value will be the square root of the original \(K_c\) value: \(K'_c = \sqrt{1.08 \times 10^7}\) \(K'_c \approx 3.28 \times 10^3\) So, the new value of \(K'_c\) is approximately \(3.28 \times 10^3\).

Calculate the new K_p value

The change in the number of moles of gas, \(\Delta n'\), for the modified balanced chemical equation is: \(\Delta n' = 1 - 1.5 = -0.5\) Using the new \(K'_c\) value and the modified \(\Delta n'\) value, calculate the new \(K'_p\) value: \(K'_p = K'_c(RT)^{\Delta{ n'}}\) \(K'_p = (3.28 \times 10^{3})(8.314)(973.15)^{-0.5}\) Solve for \(K'_p\): \(K'_p \approx 3.55 \times 10^2\) So, the new value of \(K'_p\) for the modified balanced chemical equation is approximately \(3.55 \times 10^2\).

Key concepts

Equilibrium Constants

When dealing with chemical equilibria, the equilibrium constant is a crucial element to understand. It is represented by either \(K_c\) or \(K_p\), depending on the parameters used to describe the equilibrium.

• \(K_c\) is the equilibrium constant in terms of concentration. It is used when the reactants and products are measured in molarity (mol/L).
• \(K_p\) is the equilibrium constant for reactions involving gases, given in terms of pressure.

The relationship between \(K_c\) and \(K_p\) is determined by the change in the number of moles, \(\Delta n\), of the gas particles during the reaction:

\[ K_p = K_c (RT)^{\Delta{n}} \]
where:

• \(R\) is the ideal gas constant (8.314 J/mol·K),
• \(T\) is the temperature in Kelvin,
• and \(\Delta n = n_{products} - n_{reactants} \).

In our exercise, since \(\Delta n\) is negative, it indicates a decrease in total moles of gas from reactants to products.

Gas Reactions

Gas reactions involve reactants and products that are gases. These reactions are particularly affected by changes in pressure, volume, and temperature. One feature of gas reactions is that we can use pressure instead of concentration to describe the composition of gases at equilibrium.

Gas reactions generally have equations that demonstrate how molecules interact and transform over time. In the exercise provided, the reaction is:
\[ 2 \mathrm{H}_2(g) + \mathrm{S}_2(g) \rightleftharpoons 2 \mathrm{H}_2 \mathrm{S}(g) \]
This equation tells us the stoichiometry, meaning that two moles of hydrogen gas react with one mole of sulfur gas to produce two moles of hydrogen sulfide gas. The stoichiometry is crucial in determining how the equilibrium constants relate to each reaction component, particularly when calculating the equilibrium position using \(K_c\) or \(K_p\).

A deep understanding of how \(\Delta n\) changes can affect \(K_p\) is essential, as changes in the number of gas moles affect the equilibrium in pressure-based constants, and ultimately the reactant and product concentrations at equilibrium.

Ideal Gas Law

The Ideal Gas Law is an essential principle in calculating gas-based equilibrium constants. It provides a relationship between pressure, volume, temperature, and moles of a gas:
\[ PV = nRT \]
Where:

• \(P\) is the pressure,
• \(V\) is the volume,
• \(n\) is the number of moles,
• \(R\) is the gas constant,
• and \(T\) is the temperature in Kelvin.

In our scenario for equilibrium calculations, we often focus on the pressure aspect, linking it back to \(K_p\). This connection underscores why \(K_p\) is applicable for gaseous reactions. Using the Ideal Gas Law helps us transform between \(K_c\) and \(K_p\) calculations. As seen in the exercise, converting a given temperature from Celsius to Kelvin is crucial as the gas constant \(R\) uses Kelvin. Using this law aids in not only understanding pressure but also ensuring accurate equilibrium calculations by considering temperature and volume relations when working on gas equilibrium. This connection becomes evident when translating the equilibrium equations in pressure terms, showing that understanding and applying the Ideal Gas Law is foundational for grasping chemical equilibrium concepts involving gases.