Question

Reactants \(\mathrm{A}\) and \(\mathrm{B}\) are mixed, each initially at a concentration of \(1.0 \mathrm{M}\). They react to produce \(\mathrm{C}\) according to this equation: $$ 2 \mathrm{~A}+\mathrm{B} \rightleftharpoons \mathrm{C} $$ When equilibrium is established, the concentration of \(\mathrm{C}\) is found to be \(0.30 \mathrm{M}\). Calculate the value of \(K_{\mathrm{eq}}\).

Short answer

\text{K}_{\text{eq}} \thickapprox 2.68.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation is given as: $$ 2 \text{A} + \text{B} \rightleftharpoons \text{C} $$

Define the initial concentrations

Initially, the concentrations of reactants \text{A} and \text{B} are each 1.0 M, and the concentration of product \text{C} is 0 M. $$ [\text{A}] = 1.0\text{ M}, \text{ } [\text{B}] = 1.0\text{ M}, \text{ } [\text{C}] = 0\text{ M} $$.

Find the change in concentration at equilibrium

At equilibrium, the concentration of \text{C} is found to be 0.30 M. Since 2 moles of \text{A} react with 1 mole of \text{B} to form 1 mole of \text{C}, the change in concentration for A is $$ -2x $$, \text{ B is $$ -x $$, and C is $$ +x $$ when $$ x = 0.30 $$ }. Thus, $$ [\text{A}] = 1.0 - 2(0.30) = 0.40 \text{M}, [\text{B}] = 1.0 - 0.30 = 0.70\text{ M} $$, and $$ [\text{C}] = 0.30 \text{M} $$.

Write the equilibrium expression

The equilibrium constant (\text{K}_{\text{eq}}) expression for the reaction is: $$ \text{K}_{\text{eq}} = \frac{[\text{C}]}{[\text{A}]^2[\text{B}]}. $$

Substitute equilibrium concentrations into expression

Substitute the equilibrium concentrations into the \text{K}_{\text{eq}} expression: $$ \text{K}_{\text{eq}} = \frac{0.30}{(0.40)^2 (0.70)}. $$

Calculate the value of \text{K}_{\text{eq}}

Calculate the value: $$ \text{K}_{\text{eq}} = \frac{0.30}{0.16*0.70} = \frac{0.30}{0.112} \text{ } \text{K}_{\text{eq}} \thickapprox 2.68 $$. Hence \text{K}_{\text{eq}} \thickapprox 2.68.

Key concepts

equilibrium constant

The equilibrium constant, represented as \(K_\text{eq}\), is a fundamental concept in the study of chemical equilibrium. This constant provides a ratio that reflects the concentrations of products to reactants at equilibrium for a particular reaction.
For our example, the chemical reaction given is: \(2 \text{A} + \text{B} \rightleftharpoons \text{C}\). The \(K_\text{eq}\) for this reaction can be calculated using the equilibrium concentrations of the substances involved.
The general formula for \(K_\text{eq}\) in this case is:

• \(K_\text{eq} = \frac{[\text{C}]}{[\tex{A}^2[\text{B}]}}\)

We can substitute the measured equilibrium concentrations into this equation to find the value of \(K_\text{eq}\). Such calculations help predict how the system behaves over time and under different conditions. Understanding the equilibrium constant is crucial for chemists to manipulate reactions in desired ways.

reactant and product concentrations

To understand chemical equilibrium, you need to grasp how reactant and product concentrations change over time until they reach a state of balance. In our example, reactants \(\text{A} \) and \(\text{B}\) mix, each initially at 1.0 M concentration:

• \([\text{A}] = 1.0 \text{ M}\), \(\text{B]} = 1.0 \text{ M}\)

As the reaction proceeds, these concentrations decrease as they form product \(\text{C}\), which we observe to be 0.30 M at equilibrium. The changes in concentration for A and B can be expressed as:

• \[ \text{Change in] [\text{A}] = -2x \text{ M}\]
• \[ \text{Change in] [\text{B}] = -x \text{ M}\]

to form product C with an increase of \(+x\), where \(x = 0.30 \text{ M}\). As a result, the equilibrium concentrations become:

• \([\text{A}] = 1.0 - 2 \times 0.30 = 0.40 \text{M}\)
• \([\text{B}] = 1.0 - 0.30 = 0.70 \text{M}\)
• \([\text{C}] = 0.30 \text{M}\)

These calculations help us understand how the concentrations of each substance change during the reaction and reach a point of equilibrium.

chemical reactions

Chemical reactions involve the transformation of one or more substances (reactants) into different substances (products). In the given reaction \(2 \text{A} + \text{B} \rightleftharpoons \text{C}\), A and B are reactants that combine in specific ratios to form product C. The arrow (\rightleftharpoons) indicates that this is a reversible reaction, meaning it can proceed in both forward and backward directions.
The reaction initially starts with 1.0 M concentrations of both A and B, and 0 M of C. As the reaction progresses, the concentrations of A and B decrease while C is formed until the system reaches equilibrium. At equilibrium, the rates of the forward and reverse reactions are equal, resulting in constant concentrations of the reactants and products.
Understanding the dynamics of chemical reactions, including the stoichiometric ratios (like 2 moles of A reacting with 1 mole of B), helps in predicting how the reaction will proceed. This is crucial in fields like industrial chemistry, where controlling reaction conditions is essential for producing desired products efficiently.