Question

At a particular temperature, \(K=3.75\) for the reaction $$\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g)$$ If all four gases had initial concentrations of \(0.800 M,\) calculate the equilibrium concentrations of the gases.

Short answer

The equilibrium concentrations of the gases are approximately: \([\mathrm{SO}_{2}]_\text{eqm} \approx 0.270\,\text{M}, [\mathrm{NO}_{2}]_\text{eqm} \approx 0.270\,\text{M}, [\mathrm{SO}_{3}]_\text{eqm} \approx 1.330\,\text{M}, [\mathrm{NO}]_\text{eqm} \approx 1.330\,\text{M}\).

Step-by-step solution

Write the equilibrium constant expression

The equilibrium constant expression (K) can be written for the given reaction as follows: \[K =\frac{[\mathrm{SO}_{3}][\mathrm{NO}]}{\left[\mathrm{SO}_{2}\right][\mathrm{NO}_{2}]}\]

Set up an ICE (initial-change-equilibrium) table

Set up an ICE table to represent the initial concentrations, change in concentrations, and equilibrium concentrations of reactants and products in the reaction. | | SO₂(g) | NO₂(g) | ⇌ | SO₃(g) | NO(g) | |--------|--------|--------|---|--------|-------| |Initial | 0.800 | 0.800 | | 0.800 | 0.800 | | Change | -x | -x | | +x | +x | |Eqmbrium| 0.800-x| 0.800-x| | 0.800+x | 0.800+x| Here, x represents the change in molar concentrations of reactants and products during the reaction until equilibrium is reached.

Plug the equilibrium concentrations into the K expression

Substitute the equilibrium concentrations from the ICE table into the equilibrium constant expression. Here, K = 3.75. \[3.75 = \frac{(0.800+x)(0.800+x)}{(0.800-x)(0.800-x)}\]

Solve for x

Solve the resulting equation for x by clearing the fractions and simplifying. First, multiply both sides by \((0.800 - x)^2\) \[3.75(0.800-x)^2 = (0.800+x)^2\] Expand both sides of the equation: \[2.25 - 6x +3.75x^2 = 0.64 + 1.6x + x^2\] Now, rearrange this equation to form a quadratic equation: \[2.15x^2 - 7.6x +1.61 = 0\] Solve the quadratic equation for x using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\] Where \(a = 2.15, b = -7.6,\) and \(c = 1.61\) We get two possible values for x: \(x_1 \approx 0.530\) and \(x_2 \approx 1.40\). However, since the maximum possible value for x is 0.800 (initial concentration of reactants), we discard \(x_2\). Therefore, \(x \approx 0.530\).

Calculate the equilibrium concentrations

Plug the value of x back into the equilibrium expressions from the ICE table: \[ [\mathrm{SO}_{2}]_\text{eqm} = 0.800-x = 0.800 - 0.530 \approx 0.270 \text{M}\] \[ [\mathrm{NO}_{2}]_\text{eqm} = 0.800-x = 0.800 - 0.530 \approx 0.270 \text{M}\] \[ [\mathrm{SO}_{3}]_\text{eqm} = 0.800+x = 0.800 + 0.530 \approx 1.330 \text{M}\] \[ [\mathrm{NO}]_\text{eqm} = 0.800+x = 0.800 + 0.530 \approx 1.330 \text{M}\] So, the equilibrium concentrations of the gases are approximately: \([\mathrm{SO}_{2}] \approx 0.270\,\text{M}, [\mathrm{NO}_{2}] \approx 0.270\,\text{M}, [\mathrm{SO}_{3}] \approx 1.330\,\text{M}, [\mathrm{NO}] \approx 1.330\,\text{M}\).

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K \), is a crucial concept in chemical equilibrium. It gives us a quantitative measure of the position of equilibrium for a chemical reaction. For a given balanced equation, the equilibrium constant is determined by the concentrations of products and reactants at equilibrium. The equation for \( K \) depends on the stoichiometry of the reaction. For example, in the reaction \( \mathrm{SO}_{2}(g) + \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) + \mathrm{NO}(g) \), the equilibrium constant expression is \[ K = \frac{[\mathrm{SO}_{3}][\mathrm{NO}]}{[\mathrm{SO}_{2}][\mathrm{NO}_{2}]} \].
The value of \( K \) reflects how far a reaction goes towards completion. A high \( K \) value indicates that products are favored at equilibrium, while a low value suggests reactants are favored. In this exercise, \( K = 3.75 \), which means the reaction somewhat favors the products.

ICE Table

An ICE table, which stands for Initial, Change, and Equilibrium, is a valuable tool used to systematically calculate the concentrations of reactants and products for a reaction at equilibrium. You begin with the initial concentrations of all substances involved. In this exercise, all four gases started with concentrations of 0.800 M.
The 'Change' row involves expressing the changes in concentration using a variable, typically \( x \). For the forward reaction (producing more products), the reactants decrease by \( x \), and the products increase by \( x \). Finally, the 'Equilibrium' row summarizes the concentrations at equilibrium by combining the initial concentrations with the changes. This table helps visualize and calculate the required parameters for equilibrium calculations.

Quadratic Equation

In complex equilibrium problems, such as when solving for \( x \) in this exercise, we often end up with a quadratic equation. A quadratic equation is a polynomial equation of the form \( ax^2 + bx + c = 0 \).
In our case, after substituting the equilibrium concentrations back into the equilibrium constant expression, we derived an equation: \( 2.15x^2 - 7.6x +1.61 = 0 \).
To solve for \( x \), we use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). This formula provides two potential solutions, and with the given constants, we obtained \( x_1 \approx 0.530 \) and \( x_2 \approx 1.40 \). We select \( x_1 \) because it's within the physically meaningful range determined by the initial conditions.

Reaction Stoichiometry

In chemical reactions, stoichiometry is the calculation of reactants and products in chemical reactions based on the balanced equation. This involves using the molar ratios derived from the balanced chemical equation to determine how much of each reactant participates and which products are formed.
For the given reaction: \( \mathrm{SO}_{2}(g) + \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) + \mathrm{NO}(g) \), the stoichiometry is 1:1:1:1, meaning for every mole of \( \mathrm{SO}_{2} \) and \( \mathrm{NO}_{2} \) consumed, one mole each of \( \mathrm{SO}_{3} \) and \( \mathrm{NO} \) are produced.
This stoichiometric relationship is fundamental when setting up the ICE table and calculating the changes in concentrations of each chemical species in a reaction that has reached equilibrium.