Problem 61 - Introductory Chemistry Atoms First

Question

The reaction \(\mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \rightleftarrows \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) is run in a 10.0-L vessel. The vessel is loaded with 1 mole of \(\mathrm{CO}\) and 3 moles of \(\mathrm{H}_{2}\). At equilibrium, the amounts are \(0.613\) mole of \(\mathrm{CO}, 1.839\) moles of \(\mathrm{H}_{2}, 0.387\) mole of \(\mathrm{CH}_{4}\), and \(0.387\) mole of \(\mathrm{H}_{2} \mathrm{O}\). What is the value of the equilibrium constant for this reaction? Describe the position of the equilibrium.

Step-by-Step Solution

Answer

The equilibrium constant (K) for the reaction is approximately 0.042. Since K is less than 1, the reaction favors the reactants, and the equilibrium is positioned more to the left.

Step 1: Finding the initial concentrations of the reactants and products

First, let's find the initial concentrations of all the species in the reaction. Since the reaction is run in a 10.0 L vessel: \[Initial\: concentration\: of\: CO = \frac{1\: mole}{10.0\: L} = 0.1 M\] \[Initial\: concentration\: of\: H_{2} = \frac{3\: moles}{10.0\: L} = 0.3 M\] Initially, there are no products present. So, the initial concentrations of CH4 and H2O are 0 M.

Step 2: Finding the change in concentrations as the reaction proceeds to equilibrium

Next, let's find out how the concentrations of the species change as the reaction proceeds to equilibrium. The reaction has a stoichiometry of 1:3:1:1, which means that 1 mole of CO reacts with 3 moles of H2 to produce 1 mole of CH4 and 1 mole of H2O. Since the equilibrium amount of CO is 0.613 moles and the initial amount is 1 mole, the change in the amount of CO is: \[Change\: in\: CO = (0.613 - 1) = -0.387\: moles\] For every mole of CO that reacted, 3 moles of H2 reacted. Therefore, the change in the amount of H2 is: \[Change\: in\: H_{2} = \frac{-0.387}{1} \times 3 = -1.161\: moles\] Similarly, for every mole of CO that reacted, 1 mole of CH4 and 1 mole of H2O were produced. Therefore, the change in the amounts of CH4 and H2O is: \[Change\: in\: CH_{4} = Change\: in\: H_{2}O = 0.387\: moles\]

Step 3: Finding the equilibrium concentrations of the species

Now, let's find the equilibrium concentrations of all the species. \[Equilibrium\: concentration\: of\: CO = \frac{0.613\: moles}{10.0\: L} = 0.0613 M\] \[Equilibrium\: concentration\: of\: H_{2} = \frac{1.839\: moles}{10.0\: L} = 0.1839 M\] \[Equilibrium\: concentration\: of\: CH_{4} = \frac{0.387\: moles}{10.0\: L} = 0.0387 M\] \[Equilibrium\: concentration\: of\: H_{2}O = \frac{0.387\: moles}{10.0\: L} = 0.0387 M\]

Step 4: Calculating the equilibrium constant (K)

Using the equilibrium concentrations, we can now calculate the equilibrium constant (K) for the reaction: \[K = \frac{[CH_4][H_2O]}{[CO][H_{2}]^3}\] \[K = \frac{(0.0387)(0.0387)}{(0.0613)(0.1839)^3} \approx 0.042\]

Step 5: Analyzing the position of the equilibrium

With the calculated equilibrium constant (K = 0.042), we can determine the position of the equilibrium. Since K is less than 1, the reaction favors the reactants, meaning that most of the CO and H2 are not converted to CH4 and H2O. The equilibrium is positioned more to the left, favoring the reactants. In conclusion, the equilibrium constant for this reaction is approximately 0.042 and the reaction favors the reactants, with equilibrium positioned more to the left.

Key Concepts

Chemical Equilibrium

Chemical equilibrium occurs in reversible reactions when the rate of the forward reaction equals the rate of the reverse reaction.
Even though the reactions continue to occur, there is no net change in the concentration of reactants and products because the rates are equal.

This state can be reached in any closed system, which simply means that no substances are added or removed while the reaction is taking place.

• A system reaches equilibrium when the concentrations of all reactants and products remain constant over time.
• It’s important to understand that at equilibrium, not necessarily the amounts of reactants and products will be equal.
• Equilibrium constants provide a numerical indicator of how a reaction proceeds or the relative amounts of reactants to products.

Understanding chemical equilibrium is essential for grasping how different factors can shift the equilibrium position affecting the concentrations of reactants and products over time.

Reaction Stoichiometry

Reaction stoichiometry is the quantitative study of reactants and products in a chemical reaction.
It simplifies the complex process of chemical reactions by using the mole ratios from balanced equations.
In the equation \( \ ext{CO}(g) + 3 \ ext{H}_{2}(g) \rightleftharpoons \ ext{CH}_{4}(g) + \ ext{H}_{2}\text{O}(g) \), stoichiometry tells us that:

• 1 mole of \(\text{CO}\) reacts with 3 moles of \(\text{H}_2\).
• This produces 1 mole of \(\text{CH}_4\) and 1 mole of \(\text{H}_2\text{O}\).

The stoichiometry of a reaction allows us to derive several important factors:

• Predict the volumes of gases involved.
• Calculate changes in concentration over time.
• Relate the concentration changes to the stoichiometric coefficients.

Understanding stoichiometry is crucial for calculating equilibrium concentrations and predicting how changes will impact equilibrium in future scenarios.

Equilibrium Concentrations

Equilibrium concentrations are the amounts of each reactant and product in a chemical reaction at equilibrium.
Once the reaction reaches equilibrium, these concentrations remain constant because the rates of the forward and reverse reactions are equal.
To find equilibrium concentrations, you need to know three things:

• The initial concentration of the reactants and products.
• The change in concentration as the system moves towards equilibrium.
• Use stoichiometry to relate these changes.

For the reaction \(\text{CO}(g) + 3 \text{H}_2(g) \rightleftharpoons \text{CH}_4(g) + \text{H}_2\text{O}(g)\):

• Initial concentrations might be calculated from given moles and volume.
• Calculate changes using the stoichiometric ratios provided in the equation.
• The equilibrium constant \( K \) can then be calculated using these equilibrium concentrations.

The value of the equilibrium constant provides insight into which side of the reaction—reactants or products—is favored at equilibrium.

Reversible Reactions

Reversible reactions are chemical reactions where the reactants form products, which in turn react to form the original reactants.
This dynamic process is represented with a double arrow \( \rightleftharpoons \).
Such reactions do not proceed only in a single direction.

• The rates of forward and reverse reactions influence the equilibrium position and concentrations.
• In reversible reactions, both reactants and products are present at equilibrium.

In our example reaction:
- Carbon monoxide \((\text{CO})\) reacts with hydrogen gas \((\text{H}_2)\) to form methane \((\text{CH}_4)\) and water vapor \((\text{H}_2\text{O})\).
- Despite being in equilibrium, the back and forth between reactants and products doesn’t stop, it appears static but molecular dynamics continue unabated.Understanding reversible reactions is key to predicting the outcomes and positions of equilibrium under various conditions. By mastering reversible reactions, you'll be better equipped to handle situations involving equilibrium adjustments, such as changes in temperature, pressure, and concentration.