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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 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.

Short Answer

Expert verified
The partial pressure of hydrogen gas at equilibrium is 1.25 atm, and the \(K_{\mathrm{p}}\) value for the reaction is approximately 0.122.

Step by step solution

01

Write a balanced chemical equation for the reaction

We are provided with the reaction: Nitrogen gas reacts with hydrogen gas to form ammonia. The balanced chemical equation for this reaction is: \[N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\]
02

Relate the initial partial pressures to the amounts of substances

We are given the initial partial pressures of N鈧 and H鈧 gases (P鈧佲個 and P鈧傗個) and the total pressure (P_Total) at equilibrium: P鈧佲個(N鈧) = 1.00 atm P鈧傗個(H鈧) = 2.00 atm P_Total = 2.00 atm
03

Calculate the change in partial pressure for each species

As the reaction proceeds, the partial pressures of the reactants (N鈧 and H鈧) decrease, while the partial pressure of the product (NH鈧) increases. Let's assume the partial pressure of nitrogen decreases by x. Then the partial pressure of hydrogen will decrease by 3x, since the stoichiometric ratio is 1:3 for N鈧:H鈧 in the balanced equation. Meanwhile, the partial pressure of ammonia will increase by 2x, as the stoichiometric ratio of N鈧:NH鈧 is 1:2. Thus, we have: 螖P鈧(N鈧) = -x 螖P鈧(H鈧) = -3x 螖P鈧(NH鈧) = 2x
04

Express the partial pressures at equilibrium

The partial pressures of all the gases at equilibrium can be expressed by adding the initial partial pressures and their corresponding changes during the reaction: P鈧(N鈧) = P鈧佲個(N鈧) + 螖P鈧(N鈧) = 1.00 - x P鈧(H鈧) = P鈧傗個(H鈧) + 螖P鈧(H鈧) = 2.00 - 3x P鈧(NH鈧) = 螖P鈧(NH鈧) = 2x
05

Use the equilibrium total pressure to find x

We are given the total pressure (P_Total) at equilibrium, which is 2.00 atm. Since the total pressure is the sum of all the partial pressures, we can write: P_Total = P鈧(N鈧) + P鈧(H鈧) + P鈧(NH鈧) Now, we can substitute the expressions for the partial pressures from Step 4 and find x. 2.00 = (1.00 - x) + (2.00 - 3x) + 2x 2.00 = 3.00 - 4x Let's solve for x: x = 0.25
06

Calculate the partial pressure of hydrogen gas at equilibrium

We can now substitute the value of x we found in Step 5 into the expression we derived for the partial pressure of hydrogen gas at equilibrium (P鈧(H鈧)): P鈧(H鈧) = 2.00 - 3x = 2.00 - 3(0.25) = 2.00 - 0.75 = 1.25 atm
07

Calculate the equilibrium constant, \(K_{p}\)

We can find the Kp value for the reaction by using the expression we derived in Step 4 for the partial pressures at equilibrium: \[ K_p = \frac{P_{\mathrm{NH}_{3}}^{2}}{P_{\mathrm{N}_{2}} P_{\mathrm{H}_{2}}^{3}} = \frac{(2x)^{2}}{(1.00-x)(2.00-3x)^{3}} \] Now, we will insert the value of x we found in Step 5 into the expression: \[ K_p = \frac{(2 \times 0.25)^{2}}{(1.00--0.25)(2.00-3 \times 0.25)^{3}} = \frac{(0.5)^{2}}{(0.75)(1.25)^{3}} \] Calculate the value of \(K_p\): \[ K_p = \frac{(0.5)^{2}}{(0.75)(1.25)^{3}} \approx 0.122 \] Hence, the partial pressure of hydrogen gas at equilibrium is 1.25 atm, and the \(K_{\mathrm{p}}\) value for the reaction is approximately 0.122.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Pressure
Understanding partial pressure is essential for grasping the behavior of gases in a mixture, especially when they react with each other, like in the synthesis of ammonia from nitrogen and hydrogen. Partial pressure refers to the pressure that a single gas in a mixture would exert if it alone occupied the entire volume of the mixture at the same temperature. It's a key concept in predicting the direction and extent of a chemical reaction taking place in the gaseous phase.

When dealing with chemical reactions involving gases, we must account for each gas's partial pressure to predict how the reaction proceeds and to ensure we have the correct stoichiometric ratio. In essence, the sum of the partial pressures of all individual gases in a mixture equals the total pressure exerted by the gas mixture. By knowing the partial pressures, we can apply the ideal gas law and calculate essential aspects of a reaction, such as the amounts of reactants and products at equilibrium.
Equilibrium Constant
In chemical reactions, the equilibrium constant, denoted as K, is a number that expresses the ratio of the concentrations of products to reactants at equilibrium. When dealing with gases, we use the notation Kp to represent the equilibrium constant in terms of partial pressures. The value of Kp provides insights into the position of equilibrium 鈥 that is, it tells us whether the products or reactants are favored in a closed system at equilibrium.

To calculate Kp for any reaction, we raise the partial pressures of the products and reactants to the power of their stoichiometric coefficients and then take their product over reactant ratio, as shown in the ammonia synthesis problem. A large Kp value indicates that the products are favored, while a small Kp value indicates that the reactants are favored. This knowledge is crucial because it impacts the yield of products in industrial processes and informs adjustments needed to optimize reactions.
Le Chatelier's Principle
Le Chatelier's principle provides a qualitative understanding of how a system at equilibrium reacts to external changes. It posits that if a dynamic equilibrium is disturbed by changing the conditions, such as concentration, pressure, or temperature, the position of equilibrium moves to counteract the change. This key concept helps chemists to control chemical reactions so that they can obtain the desired product yield.

For instance, increasing the pressure of gas reactants in a contained system often shifts the equilibrium toward the side of the reaction with fewer gas molecules. Conversely, adding more of a product may shift the equilibrium to form more reactants. Le Chatelier's principle not only helps in predicting the effects of such changes but is also instrumental in industrial applications, such as the Haber process for ammonia production, by maximizing the amount of desired product through careful control of reaction conditions.
Reaction Quotient
The reaction quotient, Q, is a snapshot measure of a system that tells us the relative amounts of products and reactants present during a reaction at any point in time. It is calculated in the same way as the equilibrium constant (using concentrations or partial pressures), but it is not limited to equilibrium conditions. Comparing Q with the equilibrium constant, K, allows us to predict the direction in which the reaction will proceed to reach equilibrium.

If Q is less than K, the reaction will proceed forward to produce more products. If Q is greater than K, the reaction shifts to produce more reactants. If Q equals K, the system is at equilibrium. This concept helps us understand the progression of reactions over time and to determine the necessary changes to reach equilibrium, if it's not already established.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. By understanding stoichiometry, chemists can predict the amounts of substances consumed and produced in a reaction, which allows for the precise preparation of mixtures and optimizes yields. The balanced chemical equation is the roadmap for stoichiometry; it provides the ratios in which compounds react and form products.

In our exercise, the stoichiometry of the reaction dictated that one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia. These stoichiometric coefficients were used to relate changes in partial pressures and calculate the equilibrium constant. Mastery of stoichiometry ensures efficient use of reactants, thus reducing waste and cost in chemical production.

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Most popular questions from this chapter

At a given temperature, \(K=1.3 \times 10^{-2}\) for the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\).Calculate values of \(K\) for the following reactions at this temperature. a. \(\frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\). b. \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) c. \(\mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)\). d. \(2 \mathrm{N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)\).

A \(1.00-\mathrm{L}\) flask was filled with 2.00 moles of gaseous \(\mathrm{SO}_{2}\) and 2.00 moles of gaseous \(\mathrm{NO}_{2}\) and heated. After equilibrium was reached, it was found that 1.30 moles of gaseous NO was present. Assume that the reaction $$\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g)$$, occurs under these conditions. Calculate the value of the equilibrium constant, \(K\), for this reaction.

Write the equilibrium expression \((K)\) for each of the following gas-phase reactions. a. \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)\). b. \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\). c. \(\operatorname{SiH}_{4}(g)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SiCl}_{4}(g)+2 \mathrm{H}_{2}(g)\). d. \(2 \mathrm{PBr}_{3}(g)+3 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{PCl}_{3}(g)+3 \mathrm{Br}_{2}(g)\).

Consider the decomposition equilibrium for dinitrogen pentoxide: $$2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 4 \mathrm{NO}_{2(g)+\mathrm{O}_{2}(g)$$.At a certain temperature and a total pressure of 1.00 atm, the \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(0.50 \%\) decomposed (by moles) at equilibrium. a. If the volume is increased by a factor of \(10.0,\) will the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposed at equilibrium be greater than, less than, or equal to 0.50\%? Explain your answer. b. Calculate the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) that will be decomposed at equilibrium if the volume is increased by a factor of 10.0

Novelty devices for predicting rain contain cobalt(II) chloride and are based on the following equilibrium:$$\mathrm{CoCl}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(s)$$ Purple Pink.What color will such an indicator be if rain is imminent?

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