Question

For the reaction $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ at \(700^{\circ} \mathrm{C}, K_{\mathrm{c}}=0.534 .\) Calculate the number of moles of \(\mathrm{H}_{2}\) formed at equilibrium if a mixture of 0.300 mole of \(\mathrm{CO}\) and 0.300 mole of \(\mathrm{H}_{2} \mathrm{O}\) is heated to \(700^{\circ} \mathrm{C}\) in a 10.0 - \(\mathrm{L}\) container.

Short answer

The number of moles of \(\mathrm{H}_{2}\) formed at equilibrium is approximately 0.048 mol.

Step-by-step solution

Write down the balanced chemical equation and equilibrium constant expression

The balanced chemical equation is given as: \(\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2}\mathrm{O}(g)+\mathrm{CO}(g)\), and the equilibrium constant expression based on the balanced equation is: \(K_c = \frac{[\mathrm{H}_2\mathrm{O}][\mathrm{CO}]}{[\mathrm{H}_2][\mathrm{CO}_2]}\)

Set up a table for the initial number of moles, the change in moles, and the moles at equilibrium

This table is important as it will clearly lay out the information given in the problem as well as any changes that occur when the system reaches equilibrium.

Substitute the expressions for equilibrium concentrations into the equilibrium constant

Based on equilibrium rules & the given volume, we express the equilibrium concentration in moles/L, considering the change in the number of moles as 'x'. You get an equation like \(Kc = \frac{(0.3+x)(0.3-x)}{x ⋅ 10}\)

Solve the equation for 'x'

Solving the above expression for 'x' will give the number of moles of H2 formed at equilibrium.

Key concepts

Equilibrium Constant Expression

Understanding the equilibrium constant expression is pivotal in mastering chemical equilibrium calculations. It's a mathematical representation of the relationship between the concentration of products and reactants for a reversible chemical reaction at equilibrium. For the reaction

\[ \mathrm{H}_{2}(g) + \mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2}\mathrm{O}(g) + \mathrm{CO}(g) \]

at a specified temperature, in this case, 700°C, the equilibrium constant (denoted as \(K_{c}\)) is given and it quantitatively describes the extent of the reaction. The precise formula, based on the stoichiometry of the reaction, is written as:
\[ K_c = \frac{[\mathrm{H}_2\mathrm{O}][\mathrm{CO}]}{[\mathrm{H}_2][\mathrm{CO}_2]} \]
where the square brackets denote the concentration of the gases in moles per liter (Molarity, M). An important detail to remember is that the equilibrium concentrations of pure solids and pure liquids don't appear in this expression, only gases and aqueous solutions are included.

What Does \(K_c\) Indicate?

A larger \(K_{c}\) value suggests that the reaction favors the formation of products under given conditions. Conversely, a smaller \(K_{c}\) indicates that reactants are more favored at equilibrium. Comprehending this concept deeply aids in predicting the direction in which the reaction will proceed to establish equilibrium.

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry entails the quantitative relationship between reactants and products in a chemical reaction. It's crucial for predicting the amounts of products formed from given quantities of reactants and for calculating various aspects of the reactions, such as yield.

The balanced chemical equation provides the stoichiometric coefficients, which tell us in what ratio the reactants combine and the products form. For the reaction at hand:
\[ 1 \mathrm{H}_{2}(g) + 1 \mathrm{CO}_{2}(g) \rightleftharpoons 1 \mathrm{H}_{2}\mathrm{O}(g) + 1 \mathrm{CO}(g) \]
the coefficients indicate a 1:1:1:1 ratio. This means that for every mole of \(\mathrm{H}_2\) consumed, one mole of \(\mathrm{CO}_2\) is consumed, and one mole of each, \(\mathrm{H}_2\mathrm{O}\) and \(\mathrm{CO}\), is produced.

Applying Stoichiometry to Equilibrium

In our equilibrium calculations, we apply this principle to figure out how the initial moles of reactants and products will change to reach equilibrium. This step involves setting up an ICE table (Initial, Change, Equilibrium) to visually organize and manage the quantitative data of the reaction, affording us a clear path to finding the equilibrium concentrations required to solve for \(K_c\).

Reactant and Product Concentration

The concentration of reactants and products plays a vital role in the dynamics of chemical equilibria. For a given reaction at equilibrium, this concentration is measured in molarity, which is the number of moles of a substance per liter of solution. The initial concentrations are the starting point, but as a reaction progresses toward equilibrium, these values change, usually diminishing for reactants and increasing for products, until they reach a constant value reflective of equilibrium.

In solving the given problem, knowing the initial amount of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\mathrm{O}\) and recognizing that at equilibrium, the number of moles of \(\mathrm{H}_{2}\) and \(\mathrm{CO}_2\) is the same (due to the stoichiometry of the reaction), we can introduce a variable 'x' to represent the changes in concentrations as the system reaches equilibrium. Since the volume of the container is given (10.0 L), we can convert these changes into molarity and then substitute these values into the equilibrium constant expression to solve for 'x'.

Concentration's Impact on \(K_c\)

It is important to understand that while the value of \(K_c\) is dependent on the temperature, it is independent of the initial concentrations. That said, the initial concentrations will affect the time it takes to reach equilibrium and the equilibrium concentrations themselves. This thorough understanding of concentration changes is essential for accurate equilibrium calculations.