Question

Hydrogen and carbon dioxide react at a high temperature to give water and carbon monoxide. $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftarrows \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) $$ (a) Laboratory measurements at \(986^{\circ} \mathrm{C}\) show that there are 0.11 mol each of \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}\) vapor and 0.087 mol each of \(\mathrm{H}_{2}\) and \(\mathrm{CO}_{2}\) at equilibrium in a \(1.0-\mathrm{I}\) container. Calculate the equilibrium constant for the reaction at \(986^{\circ} \mathrm{C}\) (b) Suppose 0.050 mol each of \(\mathrm{H}_{2}\) and \(\mathrm{CO}_{2}\) are placed in a 2.0-L container. When equilibrium is achieved at \(986^{\circ} \mathrm{C},\) what amounts of \(\mathrm{CO}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}),\) in moles, would be present? [Use the value of \(K\) from part (a).]

Short answer

(a) K_eq = 1.60; (b) x = 0.0225 moles, amounts are 0.0225 moles each of CO and H2O.

Step-by-step solution

Identify Given Values

For part (a), we have equilibrium concentrations in a 1.0 L container: 0.11 mol of CO and H2O, and 0.087 mol of H2 and CO2. For part (b), we start with 0.050 mol each of H2 and CO2 in a 2.0 L container.

Calculate Equilibrium Concentrations

Since the container volume is 1.0 L, for part (a), the equilibrium concentrations are the same as the mol values: \[ [\mathrm{H}_2] = 0.087 \, \text{M}, \quad [\mathrm{CO}_2] = 0.087 \, \text{M}, \quad [\mathrm{H_2O}] = 0.11 \, \text{M}, \quad [\mathrm{CO}] = 0.11 \, \text{M} \].

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K \), is a fundamental part of understanding chemical equilibrium. It is calculated using the concentrations of the products and reactants at equilibrium. For a reaction:

\[ aA + bB \rightleftharpoons cC + dD \]
the equilibrium constant \( K \) is expressed as:

\[ K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]
This formula means that the equilibrium constant is the ratio of the product of the concentrations of the products raised to their stoichiometric coefficients to the product of the concentrations of the reactants raised to their stoichiometric coefficients.

In the original exercise, you calculated \( K \) using the given concentrations at equilibrium:
- \([\mathrm{H}_2] = 0.087 \, \text{M}\)- \([\mathrm{CO}_2] = 0.087 \, \text{M}\)- \([\mathrm{H_2O}] = 0.11 \, \text{M}\)- \([\mathrm{CO}] = 0.11 \, \text{M}\)
So, \( K \) is calculated as:

\[ K = \frac{[\mathrm{H_2O}][\mathrm{CO}]}{[\mathrm{H}_2][\mathrm{CO}_2]} = \frac{(0.11)(0.11)}{(0.087)(0.087)} \]
Understanding how to find this ratio allows you to predict how changes in the system affect the equilibrium position.

Reaction Quotient

The reaction quotient, \( Q \), serves as a tool to determine the direction in which a reaction mixture will proceed to achieve equilibrium. Like the equilibrium constant, \( Q \) is calculated using the concentrations of the reactants and products at any given point in time. The formula is the same as that of the equilibrium constant:

\[ Q = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]
The key difference between \( Q \) and \( K \) is that \( K \) uses equilibrium concentrations while \( Q \) can use concentrations at any time.

Comparing \( Q \) to \( K \) tells us if a system is at equilibrium:

• If \( Q = K \), the system is at equilibrium.
• If \( Q < K \), the reaction will proceed in the forward direction.
• If \( Q > K \), the reaction will proceed in the reverse direction.

In practice, calculating \( Q \) is essential during the reaction to determine where the system stands relative to equilibrium, helping in predicting shifts and adjusting variables accordingly.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept that explains how a system at equilibrium reacts to disturbances. According to this principle, if an external change is applied to a system at equilibrium, the system adjusts itself to counteract that change and re-establish equilibrium. The changes that can affect equilibrium include:

• Concentration: Adding or removing reactants/products.
• Pressure: Changes due to volume adjustments in gaseous reactions.
• Temperature: Affects reaction rates and \( K \) values.

For instance, if a reactant is added to a system, the reaction will shift right, producing more products to reduce the added stress.

In the example of our reaction:\( \mathrm{H}_2(\mathrm{g}) + \mathrm{CO}_2(\mathrm{g}) \rightleftharpoons \mathrm{H}_2 \text{O}(\mathrm{g}) + \mathrm{CO}(\mathrm{g}) \),if more \( \mathrm{H}_2 \) were added, Le Chatelier's Principle would suggest the production of more \( \mathrm{CO} \) and \( \mathrm{H}_2\mathrm{O} \) to restore equilibrium.

Understanding Le Chatelier's Principle helps predict how a reaction might change under various experimental conditions.

Mole Calculation

Mole calculation is crucial for quantifying materials in chemistry, often required when dealing with gases and reactions. One mole of any substance contains \( 6.022 \times 10^{23} \) particles (Avogadro's number). In reactions like the one given, it's important to know how to determine moles and relate them to concentrations:

- Moles = mass (g) / molar mass (g/mol)- Concentration = moles/volume (L)

For instance, in part (b) of your exercise, starting with 0.050 mol each of \( \mathrm{H}_2 \) and \( \mathrm{CO}_2 \) in a 2.0 L container, you calculate the initial concentrations:
\[ \text{Initial } [\mathrm{H}_2] = \frac{0.050}{2.0} = 0.025 \, \text{M} \]\[ \text{Initial } [\mathrm{CO}_2] = \frac{0.050}{2.0} = 0.025 \, \text{M} \]
Such calculations are foundation blocks needed to establish equilibrium concentrations and eventually find moles of each reactant and product at equilibrium. This adept understanding is vital in solving problems involving reaction shifts and stoichiometry.