Question

Given that pure solids and liquids do not appear in \(K_{\text {eq }}\) expressions, set up the ICE chart for this reaction, given the initial conditions. $$ \begin{array}{c} \mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightleftarrows \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \\ 0.0060 \mathrm{M} \quad 0.055 \mathrm{M} \end{array} $$

Short answer

Set up the ICE chart using initial concentrations: \(\begin{array}{c|c|c|c} & \mathrm{CH_4(g)} & \mathrm{O_2(g)} & \mathrm{CO_2(g)} \\ \hline \text{Initial (M)} & 0.0060 & 0.055 & 0.0 \\ \text{Change (M)} & -x & -2x & +x \\ \text{Equilibrium (M)} & 0.0060-x & 0.055-2x & x \\ \end{array}\).

Step-by-step solution

Identify Reactants and Products in the Reaction

The given reaction is \(\mathrm{CH}_4 + 2\mathrm{O}_2 \rightleftarrows \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}\). Here, only gaseous species and aqueous solutions are included in the equilibrium expressions. Thus, \(\mathrm{H}_2\mathrm{O}(\ell)\) is not included in the \(K_\text{eq}\) expression.

Define the Initial Concentrations

Based on the initial information provided, \( \mathrm{CH}_4 \) has an initial concentration of \(0.0060\,\text{M}\) and \(\mathrm{O}_2\) has \(0.055\,\text{M}\). \(\mathrm{CO}_2\) starts with \(0.0\,\text{M}\) since it is a product that hasn't formed yet at the initial moment.

Write the Balanced Equation and Set Up ICE Chart

We establish an ICE (Initial, Change, Equilibrium) chart for the gases involved: \[\begin{array}{c|c|c|c} & \mathrm{CH_4(g)} & \mathrm{O_2(g)} & \mathrm{CO_2(g)} \\hline\text{Initial (M)} & 0.0060 & 0.055 & 0.0 \\text{Change (M)} & -x & -2x & +x \\text{Equilibrium (M)} & 0.0060-x & 0.055-2x & x \\end{array}\]

Interpreting the ICE Chart

The change for \(\mathrm{CH_4}\) is \(-x\) because it is consumed. For \(\mathrm{O_2}\), the change is \(-2x\) due to its stoichiometric coefficient of 2, and for \(\mathrm{CO_2}\), it is \(+x\) since it is produced.

Write the Equilibrium Concentrations Explicitly

Now, the concentrations at equilibrium are:- \(\mathrm{CH}_4\): \(0.0060 - x\)- \(\mathrm{O}_2\): \(0.055 - 2x\)- \(\mathrm{CO}_2\): \(x\). These are used to write down the \(K_\text{eq}\) expression.

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_{eq}\), is a mathematical expression that determines the ratio of the concentration of products to reactants at chemical equilibrium.
It focuses only on the species in the reaction that are in gaseous and aqueous states.
Solids and pure liquids do not appear in the expression for \(K_{eq}\). This is because the concentration of pure solids and liquids does not change during the reaction, thus having no effect on the equilibrium state.

• In the reaction \(\mathrm{CH}_4 + 2\mathrm{O}_2 \rightleftarrows \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}\), both \(\mathrm{CH}_4\) and \(\mathrm{O}_2\) are gases and are included in the \(K_{eq}\) calculation.
• Water, \(\mathrm{H}_2\mathrm{O}(\ell)\), as a liquid, is excluded from the \(K_{eq}\) expression.

The equilibrium constant provides insight into the position of the equilibrium.
A higher value of \(K_{eq}\) suggests that products are favored, while a lower value indicates that reactants dominate. Understanding \(K_{eq}\) is essential for predicting how the reaction will respond to changes in conditions.

Stoichiometry

Stoichiometry is fundamental when analyzing a chemical reaction, as it provides the quantitative relationship between reactants and products.
It relies on a balanced chemical equation to ensure that the law of conservation of mass is respected, which means that atoms are conserved throughout the reaction.

• In the given equation \(\mathrm{CH}_4 + 2\mathrm{O}_2 \rightleftarrows \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}\), the stoichiometric coefficients are 1, 2, 1, and 2 respectively.
• These coefficients tell us that 1 mole of methane reacts with 2 moles of oxygen to form 1 mole of carbon dioxide and 2 moles of water.

When setting up an ICE chart, the stoichiometric coefficients are used to determine the changes in concentration for each species involved.
For instance, if \(x\) moles of \(\mathrm{CH}_4\) are consumed, \(2x\) moles of \(\mathrm{O}_2\) will be used, and \(x\) moles of \(\mathrm{CO}_2\) will form.
Understanding stoichiometry helps predict how the reaction progresses from the initial state to equilibrium.

Initial Concentration

Initial concentration refers to the starting concentrations of the reactants and products before the reaction has started.
It is the starting point in setting up an ICE chart, which tracks the changes to equilibrium.

• For the reaction \(\mathrm{CH}_4 + 2\mathrm{O}_2 \rightleftarrows \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}\), the initial concentrations provided are \(0.0060\,\text{M} \) for \(\mathrm{CH}_4\) and \(0.055\,\text{M} \) for \(\mathrm{O}_2\).
• Initially, the concentration of \(\mathrm{CO}_2\) is \(0.0\,\text{M} \) as it has not been formed yet.

Accurately knowing these initial concentrations is crucial as they determine how much material is available to react.
By knowing these, we can use them to calculate the changes that occur as the system approaches equilibrium, and eventually, find the equilibrium concentrations.

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the concentrations of reactants and products remain constant over time.
It occurs when the forward reaction rate equals the reverse reaction rate, meaning no net change in concentration is observed though reactions are still happening. This is known as dynamic equilibrium.

• In the equilibrium state for the reaction \(\mathrm{CH}_4 + 2\mathrm{O}_2 \rightleftarrows \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}\), the final concentrations are determined and used to calculate the equilibrium constant expression.
• The ICE chart helps visualize this transition from initial to equilibrium state, showing how the concentrations of \(\mathrm{CH}_4\), \(\mathrm{O}_2\), and \(\mathrm{CO}_2\) change as equilibrium is reached.

Understanding chemical equilibrium is crucial for predicting and controlling reaction outcomes.
It explains why some reactions appear to stop and provides insight into how slight changes to the system can shift the equilibrium position through Le Chatelier’s principle.