Question

The equilibrium constant \(K_{P}\) for the following reaction is found to be \(4.31 \times 10^{-4}\) at \(375^{\circ} \mathrm{C}\) : $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ In a certain experiment a student starts with 0.862 atm of \(\mathrm{N}_{2}\) and 0.373 atm of \(\mathrm{H}_{2}\) in a constant-volume vessel at \(375^{\circ} \mathrm{C}\). Calculate the partial pressures of all species when equilibrium is reached.

Short answer

The exact values of the partial pressures at equilibrium can't be given without solving the equation for \(x\). The methods used to solve the equation can vary depending on the difficulty of the equation and the math skills of the students. However, once \(x\) is determined, the equilibrium pressures are found by simple substitution.

Step-by-step solution

Understand the Problem and Given Information

The reaction given is \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longleftrightarrow 2 \mathrm{NH}_{3}(g)\). The equilibrium constant (\(K_{P}\)) for this reaction at 375°C is given as \(4.31 \times 10^{-4}\). The initial pressures are \[\mathrm{P}_{N_{2}}=0.862 \,atm\,\] and \[\mathrm{P}_{H_{2}}=0.373\,atm\,\]. There is initially no \(\mathrm{NH}_{3}\). The goal is to find the partial pressures of all species at equilibrium.

Write the Expression for the Equilibrium Constant (\(K_{P}\))

The expression for the equilibrium constant \(K_{P}\) in terms of the partial pressures of the components is \(K_{P} = \frac{\left(\mathrm{P}_{NH_{3}}\right)^2}{\mathrm{P}_{N_{2}}\cdot \left(\mathrm{P}_{H_{2}}\right)^3}\), where \(\mathrm{P}_{NH_{3}}\), \(\mathrm{P}_{N_{2}}\), and \(\mathrm{P}_{H_{2}}\) are the partial pressures of the \(\mathrm{NH}_{3}\), \(\mathrm{N}_{2}\), and \(\mathrm{H}_{2}\) at equilibrium, respectively.

Write the Expression for the Changes in Partial Pressures

For every mole of \(\mathrm{N}_{2}\) that reacts, three moles of \(\mathrm{H}_{2}\) react to produce two moles of \(\mathrm{NH}_{3}\). Therefore, if \(x\) is the decrease in \(\mathrm{P}_{N_{2}}\) and \(\mathrm{P}_{H_{2}}\), then the increase in \(\mathrm{P}_{NH_{3}}\) would be \(2x\). So, the equilibrium pressures can be expressed as:\(\mathrm{P'_{N_{2}}} = \mathrm{P}_{N_{2}}-x\),\(\mathrm{P'_{H_{2}}} = \mathrm{P}_{H_{2}}-3x\),\(\mathrm{P'_{NH_{3}}} = \mathrm{P}_{NH_{3}}+ 2x = 2x\). Note that the initial pressure of \(\mathrm{NH}_{3}\) is zero.

Solve for \(x\)

Substitute the equilibrium pressure terms for \(N_{2}\), \(H_{2}\), and \(NH_{3}\) in the \(K_{P}\) expression and solve for \(x\). Thus, we have \(4.31 \times 10^{-4} = \frac{(2x)^{2}}{(0.862-x)(0.373 - 3x)^3}\). Simplifying and solving this equation will give us the value of \(x\).

Find the Equilibrium Partial Pressures

Substitute the value of \(x\) in the expressions from step 3 to get the partial pressures at equilibrium. Thus, the equilibrium pressures are:\(\mathrm{P'_{N_{2}}} = \mathrm{P}_{N_{2}}-x\),\(\mathrm{P'_{H_{2}}} = \mathrm{P}_{H_{2}}-3x\),\(\mathrm{P'_{NH_{3}}} = \mathrm{P}_{NH_{3}}+ 2x = 2x\).

Key concepts

Equilibrium Constant

In chemistry, the equilibrium constant, often denoted as \(K\), is a vital concept. It tells us how the concentrations or pressures of reactants and products relate at equilibrium in a chemical reaction. For gaseous reactions, we use \(K_{P}\), which is based on the partial pressures of the gases involved.

• The equilibrium constant depends only on the specific reaction and the temperature.
• It's easy to express the value of \(K_{P}\) using the formula derived from reaction stoichiometry.

For example, consider the reaction: \[\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\]The corresponding \(K_{P}\) expression is:\[K_{P} = \frac{\left(\mathrm{P}_{NH_{3}}\right)^2}{\mathrm{P}_{N_{2}}\cdot \left(\mathrm{P}_{H_{2}}\right)^3}\]Here, \(\mathrm{P}_{NH_{3}}, \mathrm{P}_{N_{2}}, \mathrm{P}_{H_{2}}\) are the partial pressures of ammonia, nitrogen, and hydrogen respectively. By substituting the pressures at equilibrium into this expression, we can calculate the numerical value of \(K_{P}\).

Partial Pressure

Partial pressure is a term used to describe the pressure exerted by a single gas component in a mixture of gases. It helps us understand how each gas contributes to the total pressure in a system.

• Each gas in a mixture behaves independently, as if it occupies the entire volume alone.
• The sum of partial pressures of all gases equals the total pressure.

In reactions involving gases, like our reaction between nitrogen, hydrogen, and ammonia, the partial pressures are critical for calculating equilibrium conditions. If we initially have 0.862 atm of \(\mathrm{N}_{2}\) and 0.373 atm of \(\mathrm{H}_{2}\), the change in partial pressures over time as the reaction proceeds can be determined using stoichiometry and \(K_{P}\) values. As the reaction approaches equilibrium, these partial pressures adjust according to the reaction conditions.

Reaction Stoichiometry

Reaction stoichiometry involves using the relationships between reactants and products in a balanced chemical equation. It not only helps in predicting product quantities but also in understanding how changes occur in reaction systems.

• A stoichiometric coefficient indicates how many moles of a substance participate in a reaction.
• It allows for calculation of changes in concentrations or partial pressures as reactions progress.

Consider the reaction:\[\mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\]Using stoichiometry, we say:- For every 1 mole of \(\mathrm{N}_{2}\) that reacts, 3 moles of \(\mathrm{H}_{2}\) are consumed.- 2 moles of \(\mathrm{NH}_{3}\) are produced.This relationship allows us to write the changes in partial pressures such that if \(x\) moles of \(\mathrm{N}_{2}\) react, 3\(x\) moles of \(\mathrm{H}_{2}\) are consumed, resulting in an increase of 2\(x\) moles of \(\mathrm{NH}_{3}\). Solving this helps find the equilibrium state of the system.