Question

The following equilibrium pressures were observed at a certain temperature for the reaction $$\begin{array}{c}{\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)} \\ {P_{\mathrm{NH}_{3}}=3.1 \times 10^{-2} \mathrm{atm}} \\\ {P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{atm}} \\\ {P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{atm}}\end{array}$$ Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If \(P_{\mathrm{N}_{2}}=0.525 \mathrm{atm}, P_{\mathrm{NH}_{3}}=0.0167 \mathrm{atm},\) and \(P_{\mathrm{H}_{2}}=0.00761 \mathrm{atm},\) does this represent a system at equilibrium?

Short answer

When using the given equilibrium pressures, the value of the equilibrium constant \(K_p\) is found to be 37.85. Using the new pressures, the calculated \(K_p'\) value is 33.92. The two values are not equal, which means the system is not at equilibrium with the new set of pressures.

Step-by-step solution

Write down the balanced equation and the expression for Kp

The balanced equation for the reaction is given as: \( N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g) \) We can write the expression for Kp as: \[ K_{p} = \frac{[NH_3]^2}{[N_2][H_2]^3} \] where [NH3], [N2], and [H2] represent the partial pressures of each gas.

Calculate the Kp using the given equilibrium pressures

Now, let's plug in the given equilibrium pressures into the Kp expression: \[ K_p = \frac{(3.1 \times 10^{-2})^2}{(8.5 \times 10^{-1})\times (3.1 \times 10^{-3})^3}= \frac{9.61 \times 10^{-4}}{2.5380125 \times 10^{-8}} \] Now, evaluate the equilibrium constant value: \[K_p= 37.85\]

Calculate the Kp using the new pressures

Now let's plug in the new pressures of N2, H2, and NH3 into the Kp expression and calculate the new Kp value: \[K_p' = \frac{(0.0167)^2}{(0.525)(0.00761)^3} = \frac{0.00027889}{8.22298 \times 10^{-6}}\] Evaluate the new Kp value: \[K_p'= 33.92\]

Compare the two Kp values to check for equilibrium

Let's compare the two Kp values that we have calculated. The initial Kp value was 37.85, and the new Kp value is 33.92. These values are not exactly the same, which suggests that the system is not at equilibrium with the new pressures provided.

Key concepts

Chemical Equilibrium

Chemical equilibrium is the state in which the concentrations or pressures of the reactants and products remain constant over time. This doesn't mean the reaction has stopped; instead, the forward and reverse reactions occur at the same rate.

For our reaction, \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \), equilibrium means the partial pressures of \( \text{N}_2 \), \( \text{H}_2 \), and \( \text{NH}_3 \) don't change.

Equilibrium can be dynamic, with both sides of the reaction balancing perfectly. This balance is described by the equilibrium constant, \( K_p \). It's crucial because it tells us how much product is formed compared to the reactants at a given temperature.

• If \( K_p \) is large, the products are favored.
• If \( K_p \) is small, the reactants are favored.

Understanding these concepts helps us discern whether a system has reached its equilibrium state.

Partial Pressure

Partial pressures are vital in understanding gas reactions at equilibrium. Each gas in a mixture contributes to the total pressure; this contribution is its partial pressure.

In the reaction \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \), each gas's partial pressure affects the overall equilibrium.

For example, the partial pressures given are:

• \( P_{\text{NH}_3} = 3.1 \times 10^{-2} \) atm
• \( P_{\text{N}_2} = 8.5 \times 10^{-1} \) atm
• \( P_{\text{H}_2} = 3.1 \times 10^{-3} \) atm

These values are used to calculate the equilibrium constant \( K_p \).

By comparing these pressures at different times, you can detect if the reaction mixture has reached equilibrium. If not, shifts in partial pressures indicate how the system might move toward equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle helps predict how a system at equilibrium will respond to changes. It states that if a system in equilibrium is subjected to a change, the system will adjust to counteract that change.

Applying this principle allows us to predict reactions to altered conditions, such as:

• Change in concentration
• Change in pressure
• Change in temperature

In the equation \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \), altering partial pressures can shift the equilibrium position.

For instance, if \( P_{\text{NH}_3} \) is increased, the principle suggests the system will shift to produce more \( \text{N}_2 \) and \( \text{H}_2 \) to reduce this pressure change. Conversely, increasing \( P_{\text{N}_2} \) or \( P_{\text{H}_2} \) will push the reaction towards forming more \( \text{NH}_3 \). This principle is a powerful tool for predicting the behavior of chemical reactions when external conditions change.