Problem 56 - Chemistry

Question

Nitrogen gas \(\left(\mathrm{N}_{2}\right)\) reacts with hydrogen gas \(\left(\mathrm{H}_{2}\right)\) to form ammonia \(\left(\mathrm{NH}_{3}\right) .\) At \(200^{\circ} \mathrm{C}\) in a closed container, 1.00 atm of nitrogen gas is mixed with 2.00 \(\mathrm{atm}\) of hydrogen gas. At equilibrium, the total pressure is 2.00 atm. Calculate the partial pressure of hydrogen gas at equilibrium, and calculate the \(K_{\mathrm{p}}\) value for this reaction.

Step-by-Step Solution

Answer

The equilibrium partial pressures for the reaction are \(P_{N_2} = 0.33\, atm\), \(P_{H_2} = 0\, atm\), and \(P_{NH_3} = 1.34\, atm\). However, it is impossible to calculate the \(K_p\) value due to the hydrogen pressure being equal to zero, which leads to an undefined result. This indicates that either there is an error or missing information.

Step 1: Write the balanced equation for the reaction

The balanced equation for the given reaction is: \[N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\]

Step 2: Determine the initial partial pressures of the gases

At the beginning of the reaction, we have the following partial pressures: - \(P_{N_2} = 1.00 \, atm\) - \(P_{H_2} = 2.00 \, atm\) - \(P_{NH_3} = 0 \, atm\), since no ammonia has been formed yet.

Step 3: Define changes in partial pressures

Let x be the change in the partial pressure of nitrogen gas as the reaction proceeds. The change in partial pressures for each component can be summarized as follows: - \(N_2\): \(1.00 - x\, atm\) - \(H_2\): \(2.00 - 3x \, atm\), since the stoichiometry shows that for 1 mole of \(N_2\), 3 moles of \(H_2\) react - \(NH_3\): \(2x\,atm\), since for every 1 mole of \(N_2\), we form 2 moles of \(NH_3\)

Step 4: Calculate the equilibrium partial pressure of hydrogen gas

Since the total pressure at equilibrium is given as 2.00 atm, we can write the equation as follows: \[P_{N_2} + P_{H_2} + P_{NH_3}=(1.00-x)+(2.00-3x)+(2x)=2.00\] Solving for x: \[3.00-x=2.00\] \[x=1.00\,atm\] Now we can determine the equilibrium partial pressure of hydrogen gas: \[P_{H_2} = 2.00 - 3x = 2.00 - 3(1.00) = -1.00\,atm\] However, since a negative value for pressure is not physically significant, we can conclude that all the hydrogen gas has reacted and got consumed, while a small fraction of nitrogen gas is still present. So, we can revise the partial pressures of the components at equilibrium to: - \(P_{N_2} = 1.00 - 0.67 = 0.33 \, atm\) - \(P_{H_2} = 0 \, atm\) - \(P_{NH_3} = 2 \times 0.67 = 1.34 \, atm\)

Step 5: Calculate the \(K_p\) value for the reaction

Using the balanced equation and the equilibrium partial pressures, we can compute the \(K_p\) value as follows: \[K_p = \frac{P_{NH_3}^2}{P_{N_2}\cdot P_{H_2}^3} = \frac{(1.34)^2}{(0.33)\cdot(0)^3}\] Since \(P_{H_2}^3 = 0\), the formula is undefined. This indicates that we have made an error, or we have an incomplete reaction description. In real systems, it would be highly unlikely for hydrogen gas's pressure to be exactly zero. We would usually have an infinitesimal or tiny quantity present that couldn't be accurately measured. This situation should prompt us to revisiting the problem conditions or considering uncertainties. However, based on the given problem statement, we have to admit that computing \(K_p\) is not possible.

Key Concepts

Ammonia Synthesis

Ammonia synthesis refers to the chemical process by which ammonia (\(\text{NH}_3\)) is produced from nitrogen (\(\text{N}_2\)) and hydrogen (\(\text{H}_2\)) gases. This reaction is fundamentally important in industrial chemistry because ammonia is a vital component in fertilizers and other chemical products. The synthesis occurs via the balanced chemical reaction:

• \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\)

This equation indicates that one molecule of nitrogen reacts with three molecules of hydrogen to produce two molecules of ammonia. This process typically occurs at high pressure and temperature to shift the equilibrium toward the production of ammonia. Additionally, catalysts are frequently used to speed up the reaction, making it economically viable on an industrial scale. The synthesis of ammonia is considered exothermic, releasing energy, which influences how the equilibrium shifts in response to changes in temperature.

Equilibrium Constant (Kp)

The equilibrium constant, denoted as \(K_p\), is a vital aspect in understanding chemical equilibria, particularly for gases. \(K_p\) is specifically used for reactions involving gases and is expressed in terms of partial pressures. The expression depends on the balanced chemical equation and provides insight into the position of equilibrium. For the ammonia synthesis reaction, the equilibrium constant expression is given by:

• \(K_p = \frac{(P_{\text{NH}_3})^2}{(P_{\text{N}_2})(P_{\text{H}_2})^3}\)

This equation indicates that the equilibrium constant is directly dependent on the partial pressures of the reactants and products. A higher \(K_p\) value indicates that, at equilibrium, the concentration of products is relatively higher compared to reactants, meaning the equilibrium position is shifted towards products. Conversely, a lower \(K_p\) reflects a reaction where reactants are favored. \(K_p\) is temperature-dependent, and even a slight change in temperature can significantly alter its value, thus changing the equilibrium position.

Partial Pressure Calculation

Partial pressure calculation is crucial in analyzing the behavior of gases at equilibrium. When multiple gases are present in a reaction, each gas exerts its own pressure, known as partial pressure. In the context of the ammonia synthesis reaction, you must calculate how the partial pressures of gases change as the system reaches equilibrium. Initially, the reactants have the following partial pressures:

• Nitrogen: \( 1.00 \, \text{atm} \)
• Hydrogen: \( 2.00 \, \text{atm} \)

As the reaction progresses toward equilibrium, nitrogen and hydrogen pressures decrease proportionally to their stoichiometry in the reaction. When calculating the partial pressures, assume a variable, say \(x\), representing the change in partial pressure. For example:

• Nitrogen: \(1.00 - x \)
• Hydrogen: \(2.00 - 3x \)

These calculations help determine the pressure of ammonia formed and ensure that the total pressures align with equilibrium conditions. Understanding these calculations is fundamental in predicting the amount of each substance at equilibrium and hence directly influences the calculation of \(K_p\). The step-by-step analysis of partial pressures allows us to solve for equilibrium conditions and better understand the reaction dynamics.