Question

A mixture containing 3.9 moles of \(\mathrm{NO}\) and 0.88 mole of \(\mathrm{CO}_{2}\) was allowed to react in a flask at a certain temperature according to the equation: $$ \mathrm{NO}(g)+\mathrm{CO}_{2}(g) \rightleftarrows \mathrm{NO}_{2}(g)+\mathrm{CO}(g) $$ At equilibrium, 0.11 mole of \(\mathrm{CO}_{2}\) was present. Calculate the equilibrium constant \(K_{\mathrm{c}}\) of this reaction.

Short answer

The equilibrium constant, \( K_{c} \), is 1.722.

Step-by-step solution

Write the balanced equation and define changes

The balanced equation for the reaction is \( \mathrm{NO}(g) + \mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{NO}_{2}(g) + \mathrm{CO}(g) \). Initially, we have 3.9 moles of \( \mathrm{NO} \) and 0.88 moles of \( \mathrm{CO}_{2} \). If \( x \) moles react, the change in moles will be: \( -x \) for both \( \mathrm{NO} \) and \( \mathrm{CO}_{2} \), and \( +x \) for both \( \mathrm{NO}_{2} \) and \( \mathrm{CO} \).

Determine the equilibrium moles

At equilibrium, the number of moles of \( \mathrm{CO}_{2} \) is given as 0.11 moles. Therefore, \( 0.88 - x = 0.11 \), solving this gives \( x = 0.77 \). Thus, the equilibrium moles are: \( \mathrm{NO} = 3.9 - 0.77 = 3.13 \), \( \mathrm{CO}_{2} = 0.11 \), \( \mathrm{NO}_{2} = 0.77 \), and \( \mathrm{CO} = 0.77 \).

Calculate equilibrium concentrations

To find the concentration, assume the reaction occurs in a 1 L flask, hence moles equal concentrations (Molarity), giving: \( [\mathrm{NO}] = 3.13 \), \( [\mathrm{CO}_{2}] = 0.11 \), \( [\mathrm{NO}_{2}] = 0.77 \), and \( [\mathrm{CO}] = 0.77 \).

Write the equilibrium expression

The equilibrium constant expression is \( K_{c} = \frac{[\mathrm{NO}_{2}][\mathrm{CO}]}{[\mathrm{NO}][\mathrm{CO}_{2}]} \). Insert the equilibrium concentrations into this equation.

Calculate the equilibrium constant

Substitute the values into the expression to find \( K_{c} \): \[ K_{c} = \frac{(0.77)(0.77)}{(3.13)(0.11)} \]Calculate to get:\[ K_{c} = \frac{0.5929}{0.3443} = 1.722 \]

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in which the rates of the forward and reverse reactions are equal in a chemical reaction. This means there's no net change in the concentrations of reactants and products over time. In our example, the reaction between nitric oxide (\( \mathrm{NO} \)) and carbon dioxide (\( \mathrm{CO}_2 \)) forms nitrogen dioxide (\( \mathrm{NO}_2 \)) and carbon monoxide (\( \mathrm{CO} \)).
Equilibrium signifies balance in the reaction process. Although the amounts of \( \mathrm{NO} \), \( \mathrm{CO}_2 \), \( \mathrm{NO}_2 \), and \( \mathrm{CO} \) don't change at equilibrium, the chemicals continue to react. This continuous reaction is called dynamic equilibrium.
Understanding chemical equilibrium is crucial because it determines the concentrations of reactants and products, leading to the calculation of the equilibrium constant, \( K_{c} \). This constant helps predict the extent of the reaction and the potential yield of products.

Reaction Stoichiometry

In the world of chemistry, stoichiometry refers to the quantitative relationship between elements in a chemical reaction. The balanced equation for a chemical reaction provides this stoichiometric relationship and allows chemists to predict the amounts of substances consumed and produced.
In our balanced reaction \( \mathrm{NO}(g) + \mathrm{CO}_2(g) \rightleftharpoons \mathrm{NO}_2(g) + \mathrm{CO}(g) \), each molecule of \( \mathrm{NO} \) reacts with one molecule of \( \mathrm{CO}_2 \) to produce one molecule each of \( \mathrm{NO}_2 \) and \( \mathrm{CO} \). This is a 1:1:1:1 stoichiometric relationship.
Stoichiometry helps in calculating the changes in the number of moles of each substance as the reaction proceeds toward equilibrium, which further aids in determining the equilibrium concentrations.

Balanced Equation

A balanced chemical equation is essential for understanding the chemical reaction's stoichiometry. It represents the number of units of each reactant and product involved in the reaction.
For our example, the balanced equation is: - \( \mathrm{NO}(g) + \mathrm{CO}_2(g) \rightleftharpoons \mathrm{NO}_2(g) + \mathrm{CO}(g) \).
This equation showcases the law of conservation of mass, where the number of atoms for each element is the same on both sides of the equation. Balancing ensures that the reaction accurately represents the real-life process.
With the equation properly balanced, it's possible to track how the number of molecules changes as the reaction progresses, directly impacting the calculation of the equilibrium state.

Concentration Calculation

Calculating concentrations is a critical step in understanding chemical reactions at equilibrium. Concentration is measured in \( \text{Molarity} \), \( M \), which is moles per liter. In equilibrium calculations, knowing the volume of the reaction mixture is crucial.
In our exercise, we assume the reaction occurs in a 1 L flask, which makes moles equal concentrations. Thus, the equilibrium concentrations of the substances become directly related to their moles:

• \( [\mathrm{NO}] = 3.13 \)
• \( [\mathrm{CO}_2] = 0.11 \)
• \( [\mathrm{NO}_2] = 0.77 \)
• \( [\mathrm{CO}] = 0.77 \)

These concentrations allow the calculation of the equilibrium constant \( K_c \), providing insights into the reaction's behavior and directionality. Using such calculations sets the foundation for predicting product yields and optimizing conditions for chemical processes.