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For the synthesis of ammonia $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ the equilibrium constant \(K_{\mathrm{c}}\) at \(375^{\circ} \mathrm{C}\) is \(1.2 .\) Starting with \(\left[\mathrm{H}_{2}\right]_{0}=0.76 \mathrm{M},\left[\mathrm{N}_{2}\right]_{0}=0.60 \mathrm{M},\) and \(\left[\mathrm{NH}_{3}\right]_{0}=0.48 M,\) when this mixture comes to equilibrium, which gases will have increased in concentration and which will have decreased in concentration?

Short Answer

Expert verified
When the mixture comes to equilibrium, the concentrations of \(N_{2}(g)\) and \(H_{2}(g)\) will decrease while the concentration of \(NH_{3}(g)\) will increase.

Step by step solution

01

Calculating the Reaction Quotient Qc

The reaction quotient (Qc) is a measure of the relative amounts of products and reactants present during a reaction at a particular instant. It is given by: \(Q_c = \frac{[NH_{3}]^2}{[N_{2}][H_{2}]^3}\). Substitute the initial concentrations into the equation: \(Q_c = \frac{(0.48)^2}{(0.60)*(0.76)^3} = 0.43. \)
02

Comparing Qc with Kc

The given equilibrium constant (Kc) at \(375^{\circ}C\) is 1.2. Compare the reaction quotient Qc with the equilibrium constant Kc. Since \(Q_c (0.43) < K_c (1.2)\), the reaction is not at equilibrium and will shift to the right (towards greater product concentrations) to adjust and reach equilibrium.
03

Determine the changes in concentration

The reaction shifts towards the product, meaning more reactants are converted into products. As a result, the concentrations of reactants \(N_{2}(g)\) and \(H_{2}(g)\) will decrease, while the concentration of the product \(NH_{3}(g)\) will increase.

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Key Concepts

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

Reaction Quotient (Qc)
Understanding the reaction quotient, denoted as \( Q_c \), is key in determining the direction a chemical reaction will proceed to reach equilibrium. It is calculated similarly to the equilibrium constant, \( K_c \), but is applicable to any point during the reaction, not just at equilibrium.​

For a given reaction, \( Q_c \) is defined as the ratio of the concentrations of the products to the reactants, each raised to the power of their respective coefficients in the balanced equation. For the synthesis of ammonia, the equation is:
  • \( Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \)
By calculating \( Q_c \) with initial concentrations given for ammonia (NH extsubscript{3}), nitrogen (N extsubscript{2}), and hydrogen (H extsubscript{2}), it helps to predict whether the reaction will move forward to produce more products or shift backward towards reactants to reach equilibrium.​
Equilibrium Constant (Kc)
The equilibrium constant, \( K_c \), is a crucial value that indicates the ratio of product to reactant concentrations when a reaction is at equilibrium. Unlike \( Q_c \), the equilibrium constant only applies when the system is at equilibrium. It provides insight into the extent of a reaction under specific conditions.​

For the reaction synthesizing ammonia, the equilibrium constant at 375°C is given as 1.2. This value tells us that at equilibrium, the products and reactants are in a specific ratio that results in the constant value of 1.2. When compared with \( Q_c \), it helps predict the direction the reaction needs to adjust in order to reach equilibrium conditions.​

If \( Q_c < K_c \), like in the exercise, the reaction will shift to produce more products until equilibrium is achieved. Conversely, if \( Q_c > K_c \), it indicates too many products and that the reaction needs to shift towards creating more reactants.​
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in understanding how a system at equilibrium responds to disturbances. It states that if a change is made to a system at equilibrium, the system will adjust to counteract that change and reach a new equilibrium.​​

In our case of ammonia synthesis from nitrogen and hydrogen, when \( Q_c \) was found to be less than \( K_c \), the equilibrium "shift" occurs as predicted by Le Chatelier's Principle. The system attempts to increase the product concentrations by converting more reactants into ammonia, thus increasing concentration of \( NH_3 \) and lowering concentrations of \( H_2 \) and \( N_2 \).
  • This shifting towards product is a direct application of this principle.
Understanding this helps chemists to manipulate conditions to favor either products or reactants depending on the needed outcome for industrial or laboratory processes.​
Synthesis of Ammonia
The synthesis of ammonia (NH extsubscript{3}) involves combining nitrogen (N extsubscript{2}) with hydrogen (H extsubscript{2}) gas under high temperature and pressure, a reaction crucial for producing fertilizers and various chemicals. This process is known as the Haber-Bosch process.​

Ammonia synthesis is represented by the equation: \( \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \). The process is exothermic, meaning it releases heat. Managing conditions such as temperature, pressure, and concentration of reactants/products is essential in optimizing yield.​​

Utilizing principles like \( K_c \) and Le Chatelier's, chemists can influence how efficiently the reaction produces ammonia. High pressure favors the formation of ammonia due to the reduction in number of gas molecules, while lower temperatures favor product formation due to exothermic reaction dynamics. Balancing these factors can achieve maximum efficiency in industrial practices.

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Most popular questions from this chapter

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ is 4.2 at \(1650^{\circ} \mathrm{C}\). Initially \(0.80 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.80 \mathrm{~mol}\) \(\mathrm{CO}_{2}\) are injected into a \(5.0-\mathrm{L}\) flask. Calculate the concentration of each species at equilibrium.

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ is \(3.8 \times 10^{-5}\) at \(727^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the equilibrium $$ 2 \mathrm{I}(g) \rightleftharpoons \mathrm{I}_{2}(g) $$ at the same temperature.

For the reaction $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ at \(700^{\circ} \mathrm{C}, K_{\mathrm{c}}=0.534 .\) Calculate the number of moles of \(\mathrm{H}_{2}\) formed at equilibrium if a mixture of 0.300 mole of \(\mathrm{CO}\) and 0.300 mole of \(\mathrm{H}_{2} \mathrm{O}\) is heated to \(700^{\circ} \mathrm{C}\) in a 10.0 - \(\mathrm{L}\) container.

In 1899 the German chemist Ludwig Mond developed a process for purifying nickel by converting it to the volatile nickel tetracarbonyl \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) (b.p. \(=\) \(\left.42.2^{\circ} \mathrm{C}\right)\) $$ \mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) $$ (a) Describe how you can separate nickel and its solid impurities. (b) How would you recover nickel? \(\left[\Delta H_{\mathrm{f}}^{\circ}\right.\) for \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(\left.-602.9 \mathrm{kj} / \mathrm{mol} .\right]\)

Consider the reaction $$ \begin{aligned} 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) & \\ \Delta H^{\circ}=&-198.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Comment on the changes in the concentrations of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at equilibrium if we were to \((\mathrm{a})\) increase the temperature, (b) increase the pressure, (c) increase \(\mathrm{SO}_{2},\) (d) add a catalyst, (e) add helium at constant volume.

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