Problem 85 - Chemistry The Central Science

Question

When 1.50 \(\mathrm{mol} \mathrm{CO}_{2}\) and 1.50 \(\mathrm{mol} \mathrm{H}_{2}\) are placed in a 3.00 -L container at \(395^{\circ} \mathrm{C}\) , the following reaction occurs: \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) .\) If \(K_{c}=0.802\) what are the concentrations of each substance in the equilibrium mixture?

Step-by-Step Solution

Answer

The equilibrium concentrations of the substances in the reaction are approximately: CO₂: 0.052 M, H₂: 0.052 M, CO: 0.448 M, and H₂O: 0.448 M.

Step 1: Write down the balanced equation

First, write the balanced chemical equation for the reaction: \[ CO_2(g) + H_2(g) \rightleftharpoons CO(g) + H_2O(g) \]

Step 2: Determine initial concentrations

Next, determine the initial concentrations of the reactants and products using the moles and volume of the container: Initial concentration of CO₂ = \(\frac{1.50\: mol}{3.00\: L} = 0.500\: M\) Initial concentration of H₂ = \(\frac{1.50\: mol}{3.00\: L} = 0.500\: M\) Initially, no reaction has occurred, so concentrations of CO and H₂O are both 0 M.

Step 3: Set up the ICE table

Now, we'll set up an ICE (Initial, Change, Equilibrium) table for the reaction: | | CO₂ | H₂ | CO | H₂O | |-----|-----|-----|-----|-----| | I | 0.5 | 0.5 | 0 | 0 | | C | -x | -x | +x | +x | | E |0.5-x|0.5-x| x | x | Where 'I' represents the initial concentrations, 'C' represents the change in concentrations as the reaction proceeds, and 'E' represents the equilibrium concentrations.

Step 4: Write the expression for K_c

Write the expression for K_c using the equilibrium concentrations of species: \[ K_c = \frac{[CO][H_2O]}{[CO_2][H_2]} \] Given that \(K_c = 0.802\), we can substitute the equilibrium concentrations from the ICE table: \[ 0.802 = \frac{(x)(x)}{(0.5-x)(0.5-x)} \]

Step 5: Solve the equation for x

Now, we'll solve the equation for x. It's a quadratic equation, but since the value of K_c is much less than 1, we can assume that the change in concentrations is small and x is much smaller than 0.5: \[ 0.802 = \frac{x^2}{(0.5)(0.5)} \] Solve for x: \[ x^2 = 0.802(0.25) \] \[ x = \sqrt{0.2005} \approx 0.448\: M \]

Step 6: Determine the equilibrium concentrations

Use the value of x to find the equilibrium concentrations of all species: Equilibrium concentration of CO₂ = 0.5 - x ≈ \(0.5 - 0.448 \approx 0.052\: M\) Equilibrium concentration of H₂ = 0.5 - x ≈ \(0.5 - 0.448 \approx 0.052\: M\) Equilibrium concentration of CO = x ≈ \(0.448\: M\) Equilibrium concentration of H₂O = x ≈ \(0.448\: M\) So, the equilibrium concentrations are approximately: CO₂: 0.052 M H₂: 0.052 M CO: 0.448 M H₂O: 0.448 M

Key Concepts

Equilibrium Constant (Kc)

Understanding the concept of an equilibrium constant, denoted as \( K_c \), is vital in analyzing chemical reactions. It is a numerical value that reflects the ratio of the concentrations of products to the concentrations of reactants at equilibrium, each raised to the power of their respective coefficients in the balanced equation.

For a generalized reaction where \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K_c \) is given by the expression \( K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \). Here, the square brackets denote concentration in molarity (\( M \)) at equilibrium. The constant itself is unitless and provides insight into the position of the equilibrium; a high \( K_c \) indicates a reaction favoring products, while a low \( K_c \) indicates a reaction favoring reactants. In the given problem, the \( K_c \) is 0.802, suggesting a close balance between reactants and products.

ICE Table

An ICE (Initial, Change, Equilibrium) table is a structured approach to determining the equilibrium concentrations of reactants and products in a chemical reaction. It outlines the initial concentrations, the changes that occur as the system reaches equilibrium, and the final equilibrium concentrations.

The 'I' column lists the initial concentrations before any reaction has occurred. The 'C' column denotes the change in concentration in terms of an unknown variable (often 'x'), which represents the amount of reactant consumed and product formed. The 'E' column then lists expressions for the equilibrium concentrations, which are combinations of the initial concentrations and the changes. Using the ICE table is a methodical way to visualize and calculate the shifts in a reacting system and is crucial for solving equilibrium problems, as demonstrated in the provided solution.

Equilibrium Concentrations

Equilibrium concentrations are the concentrations of each species (reactants and products) in a reaction mixture when the forward and reverse reaction rates are equal, i.e., the system is in a state of dynamic balance. Determining these values is essential to understanding the extent of a reaction and its yield.

Estimation of equilibrium concentrations often involves setting up an equation based on the equilibrium constant expression, substituting the concentrations as defined in the ICE table, and solving for the unknown variable. Once the variable is found, it is substituted back into the equilibrium expressions from the ICE table to find the concentrations of all substances present at equilibrium. Knowing these amounts can provide a detailed picture of the reaction's behavior at a given temperature.

Reaction Quotient

The reaction quotient, \( Q \), serves as a gauge for the progress of a reaction, helping one to determine whether a system is at equilibrium and, if not, in which direction it needs to shift to reach equilibrium. \( Q \) is calculated using the same formula as the equilibrium constant \( K_c \), but with the current concentrations instead of the equilibrium concentrations.

If \( Q \) is less than \( K_c \), the forward reaction is favored, and the system will shift towards the products to reach equilibrium. Conversely, if \( Q \) is greater than \( K_c \), the reverse reaction is favored, and the system will shift towards the reactants. At equilibrium, \( Q \) equals \( K_c \), indicating that the forward and reverse reactions are occurring at the same rate.