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 involved in the reaction are: $[{\mathrm{SO}}_2{]}_{eq}=[{\mathrm{NO}}_2{]}_{eq}=0.185M$ and $[{\mathrm{SO}}_3{]}_{eq}=[\mathrm{NO}{]}_{eq}=1.415M$.

Step-by-step solution

Write down the balanced chemical equation

We are already given a balanced chemical equation: ${\mathrm{SO}}_2(g)+{\mathrm{NO}}_2(g)\rightleftharpoons {\mathrm{SO}}_3(g)+\mathrm{NO}(g)$

Write the K expression for the reaction

Based on the reaction, the K expression can be written as: $K=\frac{[{\mathrm{SO}}_3][\mathrm{NO}]}{[{\mathrm{SO}}_2][{\mathrm{NO}}_2]}$ where [A] represents the concentration of any species A in the mixture.

Set up the ICE table

We will set up an ICE table to represent the initial concentrations, changes, and equilibrium concentrations for each gas involved in the reaction. ``` | SO₂ | NO₂ | SO₃ | NO Initial | 0.8 | 0.8 | 0.8 | 0.8 Change | -x | -x | +x | +x Equilibrium | 0.8-x | 0.8-x | 0.8+x | 0.8+x ```

Substitute the ICE table values into the K expression

Now we will substitute the equilibrium concentrations from the ICE table into the K expression: $3.75=\frac{(0.8+x)(0.8+x)}{(0.8-x)(0.8-x)}$

Solve the equation for x

This is a quadratic equation in terms of x. To solve this equation, we can try multiplying both sides by the denominator and simplifying the equation: $3.75(0.8-x{)}^2=(0.8+x{)}^2$ Expand and simplify the equation to get: $3=5x^2-8x$ Rearrange the equation as a standard quadratic equation: $5x^2-8x-3=0$ Now, solve the equation for x using the quadratic formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ Plugging in the values, we get: $x=\frac{8\pm \sqrt{(-8{)}^2-4(5)(-3)}}{2(5)}$ Solving for x, we get two possible values: $x=-0.215$ and $x=0.615$. Since we can't have a negative change in concentration, we will use the positive value for x: $x=0.615$

Calculate the equilibrium concentrations

Finally, use the value of x to calculate the equilibrium concentrations for each gas: - ${\mathrm{SO}}_2$ and ${\mathrm{NO}}_2$: $0.8-0.615=0.185mol{L}^{-1}$ - ${\mathrm{SO}}_3$ and $\mathrm{NO}$: $0.8+0.615=1.415mol{L}^{-1}$ So, the equilibrium concentrations of the gases are: - $[{\mathrm{SO}}_2{]}_{eq}=[{\mathrm{NO}}_2{]}_{eq}=0.185M$ - $[{\mathrm{SO}}_3{]}_{eq}=[\mathrm{NO}{]}_{eq}=1.415M$

Key concepts

Equilibrium Constant (K)

In chemical equilibrium, the equilibrium constant (K) is a crucial concept for predicting the extent of a chemical reaction under a given set of conditions. When a reversible reaction reaches equilibrium, the concentrations of the products and reactants remain constant, even though the individual molecules may continue to react back and forth.
For the reaction ${\mathrm{SO}}_2(g)+{\mathrm{NO}}_2(g)\rightleftharpoons {\mathrm{SO}}_3(g)+\mathrm{NO}(g)$, the equilibrium constant expression is given by:$$K=\frac{[{\mathrm{SO}}_3][\mathrm{NO}]}{[{\mathrm{SO}}_2][{\mathrm{NO}}_2]}$$Here, each bracket $[A]$ denotes the equilibrium concentration of a particular species $A$ in the reaction mixture, measured in molarity $M$.
Equilibrium constants can help determine how much product is formed at equilibrium. A larger $K$ value (>1) indicates a reaction favoring the formation of products, whereas a smaller $K$ value (<1) means that reactants are favored.

ICE Table

The ICE table is an essential tool for systematically tracking changes in concentrations throughout a chemical reaction progressing towards equilibrium. ICE stands for Initial, Change, and Equilibrium, representing the three stages of concentration that need to be considered.
Let's break down each component:

• Initial: This refers to the starting concentrations of the reactants and products before any change occurs. In our example, the initial concentrations of all gases are 0.800 M.
• Change: As the reaction moves toward equilibrium, we denote the changes in concentrations by "$-x$" for reactants and "+x" for products. This indicates that the reactants decrease and the products increase by an amount $x$.
• Equilibrium: After the changes are applied, we write the expressions for the concentrations at equilibrium. For example, the equilibrium concentration of ${\mathrm{SO}}_2$ will be $0.8-x$ M.

The ICE table helps visualize these changes and set up equations needed to calculate the equilibrium concentrations.

Quadratic Equation

A quadratic equation often arises in the context of solving equilibrium problems, especially when dealing with the equilibrium constant (K). These equations generally take the form $a{x}^2+bx+c=0$, where $a$, $b$, and $c$ are constants derived from the equilibrium expressions.
In the exercise, we derived a quadratic equation:$$5x^2-8x-3=0$$This equation results from substituting the equilibrium concentration values from the ICE table into the K expression.
To solve a quadratic equation, we typically use the quadratic formula:$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$Substituting our values of $a=5$, $b=-8$, and $c=-3$, we solve for $x$. Choosing the correct root is crucial. Often, we select the positive solution since a negative $x$ would imply a negative concentration, which is not possible in this context.

Concentration Calculation

After solving the quadratic equation for $x$, we use its value to calculate the equilibrium concentrations of the reacting species. It involves substituting $x$ back into the expressions from the ICE table.

• The change in reactant concentration is given by $0.8-x$, representing how much the initial concentration has decreased due to reaction progression.
• For products, the change is $0.8+x$, which shows how much the concentration has increased as products form.

For this particular problem, we calculated:

• ${\mathrm{SO}}_2$ and ${\mathrm{NO}}_2$ concentrations at equilibrium, $0.8-0.615=0.185$ M.
• ${\mathrm{SO}}_3$ and $\mathrm{NO}$ concentrations at equilibrium, $0.8+0.615=1.415$ M.

These calculations confirm the changes anticipated by the ICE table and demonstrate how equilibrium affects the concentration of each species involved in the reaction.