Question

At \(2000^{\circ} \mathrm{C}\), the equilibrium constant for the reaction \(2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)\) is \(K_{c}=2.4 \times 10^{3} .\) If the initial concentration of NO is \(0.250 \mathrm{M}\), what are the equilibrium concentrations of \(\mathrm{NO}, \mathrm{N}_{2}\), and \(\mathrm{O}_{2}\) ?

Short answer

At equilibrium, the concentrations for the given reaction are approximately \(\mathrm{[NO]} \approx 0.098\:M\), \(\mathrm{[N}_{2}] \approx 0.076\:M\), and \(\mathrm{[O}_{2}] \approx 0.076\:M\).

Step-by-step solution

Write down the initial concentrations

Initially, we are given the concentration of \(\mathrm{NO}\) which is 0.250 M. The concentrations of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) are not given, so we can assume they are initially 0 M. \[\mathrm{NO} \: \: (0.250\: M) \] \[\mathrm{N}_{2} \: \: (0 \: M)\] \[\mathrm{O}_{2} \: \: (0 \: M)\]

Set up the ICE table

Using the ICE table, we will write down the changes in the concentrations as the reaction proceeds: \( %% \begin{array}{c|c} & \mathrm{[NO]} & \mathrm{[N}_{2}] & \mathrm{[O}_{2}]\\ \hline \text{Initial} & 0.250 & 0 & 0\\ \text{Change} & -2x & x & x\\ \text{Equilibrium} & 0.250-2x & x& x \end{array} %%"\ \)

Write the equilibrium constant expression

We are given \(K_c = 2.4 \times 10^3\). Then, let's write down the equilibrium constant expression using the equilibrium concentrations from the ICE table: \[K_c = \frac{\mathrm{[N}_{2}\mathrm{]}\mathrm{[O}_{2}\mathrm{]}}{\mathrm{[NO]^2}}\]

Substitute the equilibrium concentrations and solve for x

Now we substitute the equilibrium concentrations from the ICE table and solve for x: \[ 2.4 \times 10^3 = \frac{x \cdot x}{(0.250 - 2x)^2} \] Solve for x (let the solver find the positive root of the equation, since concentrations should be positive): \(x \approx 0.076\)

Find the equilibrium concentrations

Now that we have the value of x, we can plug it back into the equilibrium concentrations from the ICE table: \[\mathrm{[NO]_{eq}} = 0.250 - 2x \approx \] \[\mathrm{[N}_{2}]_{eq} = x \approx \] \[\mathrm{[O}_{2}]_{eq} = x \approx \] The equilibrium concentrations are as follows: \[\mathrm{[NO]} \approx 0.098 \: M\] \[\mathrm{[N}_{2}] \approx 0.076 \: M\] \[\mathrm{[O}_{2}] \approx 0.076 \: M\]

Key concepts

Equilibrium Constant

The equilibrium constant, represented as \(K_c\), is a vital concept in chemistry that helps us understand the balance point of a reversible chemical reaction. This constant is derived from the concentrations of the products and reactants at equilibrium.

Let's consider the reaction \(2\text{NO}(g) \rightleftharpoons \text{N}_2(g)+\text{O}_2(g)\). Here, the equilibrium constant \(K_c\) is expressed mathematically as:

• \(K_c = \frac{[\text{N}_{2}][\text{O}_{2}]}{[\text{NO}]^2}\)

Given \(K_c = 2.4 \times 10^3\), this high value suggests that the reaction favors the formation of products \(\text{N}_2\) and \(\text{O}_2\) at equilibrium.

Finding equilibrium concentrations involves using the value of \(K_c\) to determine how much of the reactants are converted into products.

ICE Table

The ICE table is a systematic approach used to keep track of concentrations or pressures of species in a chemical reaction as it moves toward equilibrium. ICE stands for Initial, Change, and Equilibrium.

For the reaction \(2\text{NO}(g) \rightleftharpoons \text{N}_2(g)+\text{O}_2(g)\), we start by filling in the Initial concentrations:

• \([\text{NO}] = 0.250\, \text{M}\)
• \([\text{N}_2] = 0\, \text{M}\)
• \([\text{O}_2] = 0\, \text{M}\)

Next, the Change row accounts for the shift in concentrations as the reaction progresses:

• \([\text{NO}]\) decreases by \(-2x\)
• \([\text{N}_2] = x\)
• \([\text{O}_2] = x\)

Finally, the Equilibrium row shows the concentrations at equilibrium:

• \([\text{NO}] = 0.250 - 2x\)
• \([\text{N}_2] = x\)
• \([\text{O}_2] = x\)

Using the ICE table helps simplify complex equilibrium problems by breaking them into simple, manageable steps.

Chemical Reaction

A chemical reaction involves the transformation of reactants into products and can reach a state of balance known as equilibrium. In our example, the decomposition of nitrogen monoxide (NO) into nitrogen (\(\text{N}_2\)) and oxygen (\(\text{O}_2\)) is described by the equation:

• \(2\text{NO}(g) \rightleftharpoons \text{N}_{2}(g)+\text{O}_{2}(g)\)

In such reactions, equilibrium is achieved when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant.

Understanding chemical reactions involves recognizing the stoichiometry, which is the ratio of molecules involved. For this reaction, \(2\) moles of \(\text{NO}\) react to form \(1\) mole of \(\text{N}_2\) and \(1\) mole of \(\text{O}_2\).

The goal is to determine the equilibrium concentrations of all species involved, which can be calculated using the equilibrium constant \(K_c\) and the ICE table method.