Question

Consider the reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftarrows \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ At a certain temperature, the equilibrium concentrations $$ \text { are }[\mathrm{NO}]=0.31 M,\left[\mathrm{H}_{2}\right]=0.16 M,\left[\mathrm{~N}_{2}\right]=0.082 M, \text { and } $$ \(\left[\mathrm{H}_{2} \mathrm{O}\right]=4.64 \mathrm{M} .\) (a) Write the equilibrium expression for the reaction. (b) Determine the value of the equilibrium constant.

Short answer

The equilibrium constant \(K_c\) is approximately 718.22.

Step-by-step solution

Write the Equilibrium Expression

First, identify the balanced chemical equation for the reaction: \[2 \text{NO}(g) + 2 \text{H}_2(g) \rightleftharpoons \text{N}_2(g) + 2 \text{H}_2O(g)\]The equilibrium expression for the reaction is derived from the equation:\[K_c = \frac{[\text{N}_2][\text{H}_2O]^2}{[\text{NO}]^2[\text{H}_2]^2}\]

Substitute Equilibrium Concentrations

Now that we have the equilibrium expression, substitute the known concentrations into this equation. The given concentrations are:- \([\text{NO}] = 0.31 \text{ M}\)- \([\text{H}_2] = 0.16 \text{ M}\)- \([\text{N}_2] = 0.082 \text{ M}\)- \([\text{H}_2O] = 4.64 \text{ M}\)Substitute these values into the equilibrium expression:\[K_c = \frac{(0.082)(4.64)^2}{(0.31)^2(0.16)^2}\]

Calculate the Equilibrium Constant

Perform the calculations using the substituted values:First, calculate the numerator:\[(0.082)(4.64)^2 = 0.082 \times 21.5296 = 1.76543\]Next, calculate the denominator:\[(0.31)^2(0.16)^2 = 0.0961 \times 0.0256 = 0.00245856\]Finally, divide the numerator by the denominator:\[K_c = \frac{1.76543}{0.00245856} \approx 718.22\]

Conclusion: Finalize the Equilibrium Constant Value

The equilibrium constant, \(K_c\), for the reaction at this temperature is approximately \(718.22\). This value indicates the extent to which the reaction favors the formation of products over reactants at equilibrium.

Key concepts

equilibrium expression

In the study of chemistry, the equilibrium expression is a crucial component when analyzing reversible chemical reactions. It provides insight into the relation of the concentrations of the reactants and products at equilibrium. For every balanced chemical equation, there corresponds an equilibrium constant, often designated as \( K_c \), which depends on the concentrations at equilibrium.
To derive this expression for a given reaction, such as \( 2 \text{NO}(g) + 2 \text{H}_2(g) \rightleftharpoons \text{N}_2(g) + 2 \text{H}_2O(g) \), you take the product of the concentrations of the reaction products, each raised to the power of their stoichiometric coefficients, and divide it by the product of the concentrations of the reactants, similarly raised.

• For the numerator, we have \([\text{N}_2][\text{H}_2O]^2\).
• For the denominator, it's \([\text{NO}]^2[\text{H}_2]^2\).

This forms the equilibrium expression \(K_c = \frac{[\text{N}_2][\text{H}_2O]^2}{[\text{NO}]^2[\text{H}_2]^2}\), which is specific to the reaction and temperature considered.

chemical reaction

A chemical reaction occurs when substances, known as reactants, are transformed into different substances, known as products. In the context of equilibrium reactions, changes are not one-way; the reactants and products can interconvert, reaching a state of balance.
The particular reaction given, \( 2 \text{NO}(g) + 2 \text{H}_2(g) \rightleftharpoons \text{N}_2(g) + 2 \text{H}_2O(g) \), exemplifies a reversible reaction. Reversible reactions can proceed in both forward and reverse directions. The double arrow \(\rightleftharpoons\) indicates that both formation of products from reactants and the reformation of reactants from products occur. At the point of equilibrium, these opposing rates are equal. Understanding the dynamics of these reactions is key to grasping how equilibrium expressions are set up.

concentration

Concentration is a measure of the amount of a given substance in a mixture or solution and is usually expressed in moles per liter (M). In the study of chemical equilibrium, knowing the concentrations of reactants and products is essential because these values are used in calculating the equilibrium constant \( K_c \).
In the provided reaction, equilibrium concentrations are specified:

• \([\text{NO}] = 0.31 \text{ M}\)
• \([\text{H}_2] = 0.16 \text{ M}\)
• \([\text{N}_2] = 0.082 \text{ M}\)
• \([\text{H}_2O] = 4.64 \text{ M}\)

These values are fitted into the equilibrium expression for a straightforward calculation of \( K_c \). This approach helps determine the state of the reaction at equilibrium concerning the concentration of products and reactants.

chemical equilibrium

Chemical equilibrium is the state at which the concentrations of reactants and products remain constant over time, as the rates of the forward and reverse reactions are equal. It embodies the concept of balance in reversible reactions.
When a chemical system reaches equilibrium, it doesn’t mean the reactants and products are present in equal amounts. Rather, the particular ratio of concentrations defined by the equilibrium constant \( K_c \) does not change.
Understanding chemical equilibrium involves recognizing that while molecular interactions continue to occur, they do so at a pace that maintains overall concentration stability. In our reaction example, \( 2 \text{NO}(g) + 2 \text{H}_2(g) \rightleftharpoons \text{N}_2(g) + 2 \text{H}_2O(g) \), equilibrium is reflected in the computed \( K_c \) value, indicating a heavy preference toward product formation given its numeric significance.