Question

At a certain temperature, the following reactions have the constants shown: $$ \begin{array}{l} \mathrm{S}(s)+\mathrm{O}_{2}(g) \rightleftarrows \mathrm{SO}_{2}(g) \quad K_{\mathrm{c}}^{\prime}=4.2 \times 10^{52} \\ 2 \mathrm{~S}(s)+3 \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g) \quad K_{\mathrm{c}}^{\prime \prime}=9.8 \times 10^{128} \end{array} $$ Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the following reaction at that temperature: $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g) $$

Short answer

The equilibrium constant \( K_{c} \) is approximately \( 5.56 \times 10^{23} \).

Step-by-step solution

Understand the Equilibrium Constants

We are given two reactions with their respective equilibrium constants: (1) \( \mathrm{S}(s)+\mathrm{O}_{2}(g) \rightleftarrows \mathrm{SO}_{2}(g) \) with \( K_{c}' = 4.2 \times 10^{52} \), and (2) \( 2 \mathrm{S}(s)+3 \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g) \) with \( K_{c}'' = 9.8 \times 10^{128} \). We need to find the equilibrium constant for the reaction \( 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g) \).

Write the Target Reaction in Terms of Given Reactions

To find the equilibrium constant \( K_{c} \) of the target reaction \( 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g) \), note that it can be obtained by reversing the reaction for \( \mathrm{SO}_{2} \) and adding it to the \( \mathrm{SO}_{3} \) forming reaction. (1) Reverse the first reaction to obtain: \( \mathrm{SO}_{2}(g) \rightleftarrows \mathrm{S}(s)+\mathrm{O}_{2}(g) \) with the inverse of \( K_{c}' \). (2) Add the reactions.

Calculate the Equilibrium Constant of the Reverse Reaction

For the reverse reaction \( \mathrm{SO}_{2}(g) \rightleftarrows \mathrm{S}(s)+\mathrm{O}_{2}(g) \), the equilibrium constant \( K_{c1} \) is the inverse of \( K_{c}' \): \[ K_{c1} = \frac{1}{K_{c}'} = \frac{1}{4.2 \times 10^{52}} \].

Combine Reactions

Add the reverse of the first reaction (multiplied by 2) to the second reaction to form the overall reaction \[ 2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g) \].

Calculate the Overall Equilibrium Constant

The equilibrium constant \( K_{c} \) for the target reaction is the product of the equilibrium constants of the modified reactions:\[ K_{c} = (K_{c1})^2 \times K_{c}'' = \left( \frac{1}{4.2 \times 10^{52}} \right)^2 \times 9.8 \times 10^{128} \].Simplifying gives \[ K_{c} = \frac{9.8 \times 10^{128}}{(4.2 \times 10^{52})^2} \].

Simplify and Compute the Final Constant

Compute:\[ K_{c} = \frac{9.8 \times 10^{128}}{17.64 \times 10^{104}} \].This gives \[ K_{c} \approx 5.56 \times 10^{23} \].

Key concepts

Chemical Reactions

Chemical reactions are transformative processes where reactants convert into products. At the surface, these transformations involve breaking and forming bonds between atoms.
The solid sulfur (\( \mathrm{S}(s) \)) and gaseous oxygen (\( \mathrm{O}_{2}(g) \)) serve as reactants in our initial reaction examples. Such processes can proceed in one direction until reactants are exhausted, or they can be reversible, reaching a state where reactants and products are continuously interconverted.

• Reactions occur due to collisions between molecules which must have sufficient energy and correct alignment.
• The rate of a reaction depends on various factors, such as temperature and concentration.

These reactions are represented by chemical equations with reactants on the left, products on the right, and the arrow indicating the process direction. In our provided reactions, arrow symbols like \( \rightleftarrows \) indicate reversibility.

Thermodynamics

Thermodynamics in chemistry deals with the flow and conversion of energy during chemical reactions. It provides insight into reaction feasibility and energy changes.
The equilibrium constant (\( K_c \)) in reactions like \( \mathrm{S}(s)+\mathrm{O}_{2}(g) \rightleftarrows \mathrm{SO}_{2}(g) \) is linked to Gibbs free energy changes.

• Gibbs free energy determines whether reactions are spontaneous. Negative changes often indicate spontaneity.
• Exothermic reactions release heat, while endothermic reactions absorb it.

Understanding thermodynamics helps predict if, and to what extent, a reaction will proceed.

Reversible Reactions

Reversible reactions are processes where the products can reform into reactants, achieving a state of equilibrium over time.
In this state, the forward and reverse reactions occur at the same rate, and concentrations of reactants and products remain constant. For example, \( 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g) \) is reversible, indicated by \( \rightleftarrows \).

• Equilibrium doesn't mean equal concentrations but rather a balance of forward and reverse reactions.
• The equilibrium position depends on the reaction's conditions, like temperature and pressure.

The concept of equilibrium is crucial in determining the extent to which reactants are converted into products, governed by their respective equilibrium constants.

Reaction Mechanisms

Reaction mechanisms describe the step-by-step pathway from reactants to products. They break down reactions into simpler stages, providing insights into the sequence and nature of elementary processes involved.
Steps involve formation and breaking of chemical bonds and include intermediates that are not seen in the overall reaction equation.

• Each step in a mechanism has its own rate, influencing the overall reaction speed.
• Catalysts can alter mechanisms, usually lowering activation energy, increasing the reaction rate.

Analyzing mechanisms helps chemists understand not just what happens in a reaction, but how it unfolds, which can lead to improvements in reaction control and efficiency.