Problem 24 - Chemistry and Chemical Reactivity

Question

Calculate \(K\) for the reaction $$ \mathrm{Fe}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{FeO}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g}) $$ given the following information: $$\begin{array}{ll} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) & K=1.6 \\ \mathrm{FeO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) & \mathrm{K}=0.67 \end{array}$$

Step-by-Step Solution

Answer

The equilibrium constant \( K \) for the reaction is approximately 2.39.

Step 1: Identify Relevant Reactions

We are given two reactions with their equilibrium constants:1. \( \mathrm{H}_{2} \mathrm{O} (\mathrm{g}) + \mathrm{CO} (\mathrm{g}) \rightleftharpoons \mathrm{H}_{2} (\mathrm{g}) + \mathrm{CO}_{2} (\mathrm{g}) \) with \( K_1 = 1.6 \)2. \( \mathrm{FeO} (\mathrm{s}) + \mathrm{CO} (\mathrm{g}) \rightleftharpoons \mathrm{Fe} (\mathrm{s}) + \mathrm{CO}_{2} (\mathrm{g}) \) with \( K_2 = 0.67 \)Our target is to find the equilibrium constant \( K \) for the reaction: \( \mathrm{Fe} (\mathrm{s}) + \mathrm{H}_{2} \mathrm{O} (\mathrm{g}) \rightleftharpoons \mathrm{FeO} (\mathrm{s}) + \mathrm{H}_{2} (\mathrm{g}) \).

Step 2: Reverse and Modify the Reactions

Reverse the second given reaction to align it closer to the target reaction:\( \mathrm{Fe} (\mathrm{s}) + \mathrm{CO}_{2} (\mathrm{g}) \rightleftharpoons \mathrm{FeO} (\mathrm{s}) + \mathrm{CO} (\mathrm{g}) \).Reversing the reaction changes its equilibrium constant to \( \frac{1}{K_2} = \frac{1}{0.67} \approx 1.493 \).

Step 3: Combine Reactions

Combine the modified second reaction with the first given reaction:1. \( \mathrm{H}_{2} \mathrm{O} (\mathrm{g}) + \mathrm{CO} (\mathrm{g}) \rightleftharpoons \mathrm{H}_{2} (\mathrm{g}) + \mathrm{CO}_{2} (\mathrm{g}) \) (\( K_1 = 1.6 \))2. \( \mathrm{Fe} (\mathrm{s}) + \mathrm{CO}_{2} (\mathrm{g}) \rightleftharpoons \mathrm{FeO} (\mathrm{s}) + \mathrm{CO} (\mathrm{g}) \) (\( K_2' \approx 1.493 \))By adding these reactions:\( \mathrm{H}_{2} \mathrm{O} (\mathrm{g}) + \mathrm{CO} (\mathrm{g}) + \mathrm{Fe} (\mathrm{s}) + \mathrm{CO}_{2} (\mathrm{g}) \rightleftharpoons \mathrm{H}_{2} (\mathrm{g}) + \mathrm{CO}_{2} (\mathrm{g}) + \mathrm{FeO} (\mathrm{s}) + \mathrm{CO} (\mathrm{g}) \)Cancel out the \( \mathrm{CO} \) and \( \mathrm{CO}_{2} \) terms as they appear on both sides to achieve the target reaction:\( \mathrm{Fe} (\mathrm{s}) + \mathrm{H}_{2} \mathrm{O} (\mathrm{g}) \rightleftharpoons \mathrm{FeO} (\mathrm{s}) + \mathrm{H}_{2} (\mathrm{g}) \).

Step 4: Calculate the Overall Equilibrium Constant

The equilibrium constant for the combined reaction is the product of the equilibrium constants from the modified reactions:\( K = K_1 \times K_2' = 1.6 \times 1.493 \approx 2.389 \).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a fascinating concept in chemistry where the rate of the forward reaction equals the rate of the reverse reaction.
This balance means that the concentrations of reactants and products remain constant over time, creating a steady state.
Even though reactions are continuously occurring, there's no net change in the quantities of substances. In our exercise, we determine the equilibrium constant (\(K\)) for the reaction:

• Fe (s) + H2O (g) \(\rightleftharpoons\) FeO (s) + H2 (g)

The equilibrium constant is a mathematical expression derived from the law of mass action.
For a general reaction like \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant is given by:\[K = \frac{{[C]^c[D]^d}}{{[A]^a[B]^b}}\]For reactions involving solids, such as our exercise, only gases and aqueous species are included in the expression. Understanding chemical equilibrium helps us comprehend how and why reactions reach a state of balance, which is key in predicting the extent of chemical reactions under controlled conditions.

Thermodynamics

Thermodynamics plays a crucial role in explaining chemical equilibrium.
It's a branch of physical science that examines how heat, work, and energy affect matter.The link between thermodynamics and equilibrium lies mainly in Gibbs free energy (\(\Delta G\)).
For a reaction at constant temperature and pressure, the change in Gibbs free energy must be zero at equilibrium:\[\Delta G = 0\]This condition tells us how energy is distributed in a system, influencing whether a reaction is spontaneous or not.
A negative \(\Delta G\) implies a spontaneous process, while positivity means non-spontaneous.The equilibrium constant \(K\) is also related to Gibbs free energy through the equation:\[\Delta G = -RT \ln K\]Where \(R\) is the gas constant and \(T\) is the temperature in Kelvin.
This correlation helps predict how changes in temperature can shift equilibrium, demonstrating the fundamental connection between thermodynamics and chemical reactions.

Reversible Reactions

Reversible reactions are types of chemical reactions that can proceed in both the forward and reverse directions.
In equilibrium, both processes happen simultaneously, but there isn't any overall change over time.From the exercise, we learned about two separate reversible reactions:

• \(\mathrm{H}_{2} \mathrm{O} (g) + \mathrm{CO} (g) \rightleftharpoons \mathrm{H}_{2} (g) + \mathrm{CO}_{2} (g)\)
• \(\mathrm{FeO} (s) + \mathrm{CO} (g) \rightleftharpoons \mathrm{Fe} (s) + \mathrm{CO}_{2} (g)\)

Both reactions have their own equilibrium constants which illustrate their respective states of balance.Reversibility in reactions is vital in industrial processes and biochemical pathways.
For example, in catalytic converters in cars or the synthesis of ammonia in the Haber process.
It highlights the dynamic nature of reactions, where both products and reactants are interconvertible based on conditions like temperature and pressure.Understanding reversible reactions is essential as it showcases the shifting nature of chemical processes, allowing us to grasp how conditions influence reaction directions on a molecular level.