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?

Short answer

The equilibrium concentrations of the species are: CO2: \( 0.184 \thinspace M \), H2: \( 0.184 \thinspace M \), CO: \( 0.316 \thinspace M \), and H2O: \( 0.316 \thinspace M \).

Step-by-step solution

Calculate initial concentrations

Calculate the initial concentrations of the reactants CO2 and H2, and assume that both products, CO and H2O, have an initial concentration of 0 mol/L since the reaction has not yet started. To do this, divide the given moles by the volume of the container (3.00 L). Initial concentrations: \[ [CO_2] = \frac{1.50 \thinspace mol}{3.00 \thinspace L} = 0.50 \thinspace M \] \[ [H_2] = \frac{1.50 \thinspace mol}{3.00 \thinspace L} = 0.50 \thinspace M \] \[ [CO] = [H_2O] = 0 \thinspace M \]

Set up the ICE table

Construct an ICE (Initial, Change, Equilibrium) table to show the changes in concentrations as the reaction reaches equilibrium. | | CO2 | + | H2 | ⇌ | CO | + | H2O | |---------------|-----|---|----|----|----|---|-----| | Initial (M) | 0.5 | | 0.5 | | 0 | | 0 | | Change (M) | -x | | -x | | +x | | +x | | Equilibrium |0.5-x| |0.5-x| | x | | x |

Write the equilibrium expression

Write the equilibrium expression (Kc) for the given reaction, and plug the equilibrium concentrations from the ICE table into the expression. \[ K_c = \frac{[CO][H_2O]}{[CO_2][H_2]} \] Given Kc = 0.802: \[ 0.802 = \frac{(x)(x)}{(0.5-x)(0.5-x)} \]

Solve for x

Solve the equation for x, which represents the change in concentration for each species as the reaction reaches equilibrium. Notice that this equation is a quadratic equation in the form of \( Ax^2 + Bx + C = 0 \). To solve the quadratic equation, you can use the quadratic formula: \[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \] However, in this case, it is reasonable to assume \( x \ll 0.5 \), so we can simplify the equation as follows: \[ 0.802 = \frac{x^2}{(0.5)(0.5)} \] Now solve for x: \[ x^2 = 0.802 \times (0.5)(0.5) \] \[ x = \sqrt{0.802 \times (0.5)(0.5)} \] \[ x \approx 0.316 \]

Calculate equilibrium concentrations

Now that we have the value of x, find the equilibrium concentrations of all the species by plugging x back into the values from the ICE table: \[ [CO_2]_{eq} = 0.5 - x = 0.5 - 0.316 \approx 0.184 \thinspace M \] \[ [H_2]_{eq} = 0.5 - x = 0.5 - 0.316 \approx 0.184 \thinspace M \] \[ [CO]_{eq} = x \approx 0.316 \thinspace M \] \[ [H_2O]_{eq} = x \approx 0.316 \thinspace M \] Equilibrium concentrations: CO2: 0.184 M H2: 0.184 M CO: 0.316 M H2O: 0.316 M

Key concepts

ICE Table

An ICE table is a useful tool for solving equilibrium problems in chemistry. The acronym stands for Initial, Change, and Equilibrium, which are the stages of concentration values during a chemical reaction as it approaches equilibrium.
Firstly, list the initial concentrations. These are the starting concentrations of your reactants and products before any reaction occurs. For our example,

• CO2 and H2 both start at 0.5 M,
• CO and H2O begin at 0 M since they haven't been produced yet.

The 'Change' row in the ICE table describes how concentrations shift as the reaction proceeds toward equilibrium.

• Reactants decrease by a value of x (hence, -x),
• Products increase by the same amount (hence, +x) since they are formed from the reactants.

Finally, the 'Equilibrium' row shows the concentrations of all species when the system has reached equilibrium. Here, we express the reactant concentrations as 0.5 - x, while the products are simply x.
The ICE table is crucial because it systematically tracks concentration changes, setting the stage for calculating equilibrium concentrations using mathematical expressions.

Equilibrium Constant (Kc)

The equilibrium constant, denoted as \(K_c\), measures the ratio of concentrations of products to reactants at equilibrium for a given reaction. It is specific for each reaction and depends on temperature, but not on the initial concentrations of reactants and products.
In a typical reaction of the form \(aA + bB \rightleftharpoons cC + dD\), the expression for \(K_c\) is written as:

• \(K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\)

For our reaction \(\mathrm{CO}_2(g)+\mathrm{H}_2(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_2\mathrm{O}(g)\), the equilibrium expression is:

• \(K_c = \frac{[\mathrm{CO}][\mathrm{H}_2\mathrm{O}]}{[\mathrm{CO}_2][\mathrm{H}_2]}\)

We know that \(K_c = 0.802\), indicating that at equilibrium, the system favors products slightly more than reactants. By substituting the equilibrium concentrations from the ICE table into this expression, we are able to solve for x, the change in concentration that occurs upon reaching equilibrium.

Quadratic Equation in Chemistry

The appearance of a quadratic equation in equilibrium problems is common, as equilibrium expressions often involve ratios that create such equations. Once equilibrium concentrations are expressed with variables like x, we frequently encounter quadratic equations of the form \(Ax^2 + Bx + C = 0\).
To solve a quadratic equation, the quadratic formula is used:\(x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\)
In this exercise, we assume \(x \ll 0.5\) for simplification, considering a small shift in concentration relative to initial values. This approach allows us to skip solving the full quadratic equation directly. Instead, we simplify to \(0.802 = \frac{x^2}{(0.5)(0.5)}\) and solve for x directly.
When faced with a more complex equilibrium problem, using the quadratic formula helps find the precise x-value. This demonstrates the relationship and interplay between mathematical techniques and chemical processes.