Question

For the reaction $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g) $$ it is found at equilibrium at a certain temperature that the concentrations are \([\mathrm{CO}(g)]=2.7 \times 10^{-4} M,\left[\mathrm{O}_{2}(g)\right]=1.9 \times 10^{-3} M,\) and \(\left[\mathrm{CO}_{2}(g)\right]=1.1 \times 10^{-1} M\) Calculate \(K\) for the reaction at this temperature.

Short answer

The equilibrium constant (K) for the given reaction at this temperature is approximately 4.29.

Step-by-step solution

1. Write the balanced reaction

The balanced reaction is already given: \[ 2 CO(g) + O_2(g) \rightleftharpoons 2 CO_2(g) \]

2. Write the equilibrium expression

For the given balanced reaction, the equilibrium expression can be written as the ratio of the concentrations of products to reactants, each raised to the power of their coefficients in the balanced equation. So, K is: \[ K = \frac{[CO_2]^2}{[CO]^2 \times [O_2]} \]

3. Substitute equilibrium concentrations into the expression

The given concentrations at equilibrium are: - [CO] = \(2.7 \times 10^{-4}\) M - [O2] = \(1.9 \times 10^{-3}\) M - [CO2] = \(1.1 \times 10^{-1}\) M Now, we will substitute the given concentrations into the equilibrium expression: \[ K = \frac{(1.1 \times 10^{-1})^2}{(2.7 \times 10^{-4})^2 \times (1.9 \times 10^{-3})} \]

4. Calculate the equilibrium constant

Perform the calculations in the equation to find the value of K: \[ K \approx \frac{(1.1 \times 10^{-1})^2}{(2.7 \times 10^{-4})^2 \times (1.9 \times 10^{-3})} \approx 4.288 \] So, the equilibrium constant (K) for the given reaction at this temperature is approximately 4.29.

Key concepts

Understanding Chemical Equilibrium

Chemical equilibrium is a state in which a chemical reaction and its reverse reaction occur at the same rate, which leads to a constant concentration of reactants and products. It's very much like a tug-of-war where both teams are equally strong; nobody wins, but the effort continues. In chemistry, this happens when the rate at which the reactants produce products is equal to the rate at which products change back into reactants.

Imagine you've got a canister of carbon monoxide and oxygen reacting to form carbon dioxide in a closed system. Over time, the amount of carbon dioxide increases as carbon monoxide and oxygen decrease. But at a certain point, the carbon dioxide starts to decompose back into carbon monoxide and oxygen at the same rate it is being formed. The system has reached chemical equilibrium. The amount of each substance remains constant, although reactions continue to occur on the molecular level.

This concept is vital to understand as it applies to many reactions in nature and industry, determining the composition of the mixture at any given time.

Equilibrium Expressions and the Equilibrium Constant (K)

An equilibrium expression provides a way to quantify the position of equilibrium for a reversible chemical reaction. It's a mathematical representation of the concentrations of the products divided by the concentrations of the reactants. Each concentration is raised to the power of its coefficient from the balanced chemical equation.

For the reaction \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)\), we can denote the equilibrium constant (\(K\)) expression as:\[K = \frac{[\mathrm{CO}_2]^2}{[\mathrm{CO}]^2 \times [\mathrm{O}_2]}\]

The square brackets indicate the concentration of each species, and the superscripts correspond to the coefficient of that species in the balanced equation. This particular value of \(K\) tells us about the extent of the reaction; a larger \(K\) indicates that at equilibrium, the concentration of the products is much higher than that of the reactants, favoring product formation.

Calculating Equilibrium Concentrations

When dealing with equilibrium concentrations, it's important to plug in the actual concentrations measured at equilibrium into the equilibrium expression. The exercise provided shows the concentrations of carbon monoxide, oxygen, and carbon dioxide at equilibrium, which can be substituted directly into the expression to calculate the equilibrium constant, \(K\).

Let's recall our equilibrium expression:\[K = \frac{[\mathrm{CO}_2]^2}{[\mathrm{CO}]^2 \times [\mathrm{O}_2]}\]Now, we fill in the given concentrations to find \(K\):\[K = \frac{(1.1 \times 10^{-1})^2}{(2.7 \times 10^{-4})^2 \times (1.9 \times 10^{-3})} \approx 4.29\]

By solving this, we can understand the system at equilibrium and predict the behavior of the reaction mixture. This calculation informs us that under the given conditions, the reaction favors the production of carbon dioxide. Knowing how to manipulate such equations not only helps us predict the outcome of reactions but also plays a crucial role in industrial processes, where achieving a desired balance of reactants and products is essential for cost-efficiency and quality control.