Problem 103 - Chemistry An Atoms First Approach

Question

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}}=5.3 \times 10^{5}\) for the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ When a certain partial pressure of \(\mathrm{NH}_{3}(g)\) is put into an otherwise empty rigid vessel at \(25^{\circ} \mathrm{C},\) equilibrium is reached when \(50.0 \%\) of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred?

Step-by-Step Solution

Answer

The original partial pressure of ammonia (P₀) before any decomposition occurred was 8 atm.

Step 1: Write the equilibrium constant expression

For the given chemical equation, the equilibrium constant expression is: \[K_p = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2] [\mathrm{H}_2]^3}\]

Step 2: Set up an ICE table

We can represent the changes in partial pressures using an ICE table. Here, "I" stands for initial, "C" for change, and "E" for equilibrium: ``` N2 (g) | 3H2 (g) | 2NH3 (g) --------------------------------- I | 0 | 0 | P₀ C | +x | +3x | -2x E | x | 3x | P₀-2x ```

Step 3: Substitute the values into Kp expression

We are given that 50% of the original NH3 decomposes. Therefore, x = P₀ / 2, and we can substitute these values for N2, H2, and NH3 into our Kp expression: \[K_p = \frac{(P₀-2x)^2}{(x)(3x)^3} = \frac{(P₀ - P₀)}{(P₀/2)(3P₀/2)^3}\]

Step 4: Solve for the original partial pressure of ammonia (P₀)

We are given that \(K_p = 5.3 \times 10^5\). Substitute this value for Kp and solve for P₀: \[5.3 \times 10^5 = \frac{(P₀ - P₀)^2}{(P₀/2)(3P₀/2)^3}\] Cross multiply and simplify the equation: \[5.3 \times 10^5 \cdot (P₀/2)(3P₀/2)^3 = (P₀ - P₀)^2\] \[(5.3 \times 10^5) \cdot (\frac{27P₀^4}{8}) = P₀^2 - 2P₀^2 + P₀^3\] Divide both sides by \(27/8\) and rearrange the equation to form a cubic equation: \[P₀^3 - 8P₀^2 + \frac{8 \times (5.3 \times 10^5)}{27}P₀ = 0\] Now, we can either solve the cubic equation analytically or use numerical methods. In this case, we can observe that P₀ = 8 is a solution for the equation. Now we know one root of the equation and can further simplify: \[(P₀ - 8)(P₀^2 - 2P₀ + \frac{8 \times (5.3 \times 10^5)}{27}) = 0\] Since partial pressure cannot be negative, the value of P₀ should be positive. Thus, we can ignore the quadratic factor in the equation above and consider P₀=8. So, the original partial pressure of NH3 (P₀) was 8 atm.

Key Concepts

Equilibrium Constant Expression

The equilibrium constant expression for a chemical reaction quantifies the ratio of product concentrations to reactant concentrations at equilibrium. Each concentration is raised to the power of its respective coefficient in the balanced equation.

For example, for the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2\mathrm{NH}_{3}(g)\), the equilibrium constant expression in terms of partial pressures (\(K_p\)) is represented as:\[K_p = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2] [\mathrm{H}_2]^3}\]
The equilibrium constant, \(K_p\), is determined experimentally and remains constant at a given temperature, exemplifying the relationship between reactants and products at equilibrium.

ICE Table

An ICE (Initial, Change, Equilibrium) table is a systematic way to organize and calculate the concentrations or partial pressures of reactants and products through the course of a reaction reaching equilibrium.

Initially, the table lists the starting conditions. 'Change' denotes the shift in concentrations or pressures as the system moves towards equilibrium. 'Equilibrium' values are those when the system is at rest. It's essential for solving problems like the decomposition of ammonia, where one must determine the shift in concentrations to reach the equilibrium state.

Partial Pressures

In gas-phase reactions, the partial pressure of each gas is an essential component of the equilibrium state. Partial pressure is the pressure that a single component in a mixture of gases would exert if it alone occupied the entire volume.

In our exercise, these are used within the ICE table to track how the pressure of ammonia and the produced nitrogen and hydrogen gases change as the system reaches equilibrium. Understanding partial pressures helps predict how the reaction will shift to maintain equilibrium when external conditions are manipulated.

Reaction Quotient

The reaction quotient (\(Q\)) serves as a 'snapshot' of a reaction’s status. Calculated using the same formula as the equilibrium constant, it can be used at any point during the reaction to determine the direction in which the reaction mixture will shift to reach equilibrium.

If \(Q < K\), the system will shift to produce more products. Conversely, if \(Q > K\), the reaction will shift towards the reactants. When \(Q = K\), the system is at equilibrium. This guides us in predicting the behavior of a reaction before it arrives at its equilibrium state.

Le Chatelier's Principle

Le Chatelier's Principle provides insight into how a system at equilibrium responds to changes in concentration, pressure, volume, or temperature. According to this principle, if an external change is applied to a system at equilibrium, the system adjusts in a way that seeks to counteract the change.

This principle is widely used to predict the effects of such changes. For instance, increasing the pressure of the system by reducing the volume would shift the equilibrium towards the side with fewer moles of gas. It is a key concept when considering how to alter conditions to favor the formation of desired products.