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A reaction vessel contains \(\mathrm{NH}_{3}, \mathrm{~N}_{2}\), and \(\mathrm{H}_{2}\) at equilibrium at a certain temperature. The equilibrium concentrations are \(\left[\mathrm{NH}_{3}\right]=0.25 M,\left[\mathrm{~N}_{2}\right]=0.11 M,\) and \(\left[\mathrm{H}_{2}\right]=1.91 M\) Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the synthesis of ammonia if the reaction is represented as: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftarrows 2 \mathrm{NH}_{3}(g)\) (b) \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftarrows \mathrm{NH}_{3}(g)\)

Short Answer

Expert verified
(a) \(K_c \approx 0.0815\); (b) \(K_c' \approx 0.285\).

Step by step solution

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01

Identify the Reaction Equation

First, we identify the given chemical reactions for which we need to calculate the equilibrium constant \(K_c\). We have two reactions: (a) \(\mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightleftarrows 2\mathrm{NH}_3(g)\) and (b) \(\frac{1}{2}\mathrm{N}_2(g) + \frac{3}{2}\mathrm{H}_2(g) \rightleftarrows \mathrm{NH}_3(g)\).
02

Write the Expression for Equilibrium Constant, Kc

The equilibrium constant expression \(K_c\) for a generic reaction \(aA + bB \rightleftarrows cC + dD\) is given by:\[ K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]
03

Calculate Kc for the First Reaction

For reaction (a), \(\mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightleftarrows 2\mathrm{NH}_3(g)\), we can write the equilibrium constant \(K_c\) as:\[ K_c = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} \]Substitute the given concentrations: \[ K_c = \frac{(0.25)^2}{(0.11)(1.91)^3} \]Calculate the values in the numerator and denominator to find \(K_c\).
04

Carry out the Actual Calculation of Kc

Calculate each part of the equation:- Numerator: \( (0.25)^2 = 0.0625 \)- Denominator: \( (0.11) \times (1.91)^3 = 0.11 \times 6.9771 = 0.767 \)Thus, \[ K_c = \frac{0.0625}{0.767} \approx 0.0815 \]
05

Calculate Kc for the Second Reaction

For reaction (b), \(\frac{1}{2}\mathrm{N}_2(g) + \frac{3}{2}\mathrm{H}_2(g) \rightleftarrows \mathrm{NH}_3(g)\), the equilibrium constant \(K_c'\) is related to the first reaction by: Since the second reaction is half of the first reaction, \(K_c' = \sqrt{K_c}\).Calculate \(K_c'\) using the value obtained from step 4:\[ K_c' = \sqrt{0.0815} \approx 0.285 \]
06

Confirm the Units and Final Result

Check your results and ensure they are dimensionless as per equilibrium constant conventions in concentration terms. Conclude with the calculated values for both reactions.

Key Concepts

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

Chemical Equilibrium
Chemical equilibrium refers to a state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time. Though the molecules continue to react, there is no net change in their concentrations.

This state occurs when the reactions occur at the same rate in both directions. Therefore, equilibrium is dynamic, not static. When explaining this concept, imagine two opposing relay runners passing a baton back and forth at the same speed. In chemical reactions, the molecules are like these runners exchanging places without a change in the overall number in each location.

The significance of achieving equilibrium is critical in predicting how reactions proceed under different conditions. It allows chemists to understand how various factors, like temperature and pressure, affect the reaction dynamics.
Concentration
The concentration of a substance in a solution refers to the amount of solute present in a given volume of solvent. In chemical equilibrium, concentration is a key factor because it directly influences the position of the equilibrium.

In our specific example of ammonia synthesis, we have given concentrations of
  • Ammonia ( ext{NH}_3) at 0.25 M,
  • Nitrogen ( ext{N}_2) at 0.11 M, and
  • Hydrogen ( ext{H}_2) at 1.91 M.
These concentrations are used to determine the equilibrium constant, as they represent the concentrations at equilibrium.

When the concentration of a reactant or product increases or decreases, it can shift the equilibrium position to restore balance, according to Le Chatelier's principle. Thus, understanding concentration helps predict the changes in a reaction's behavior under different scenarios.
Equilibrium Expression
In chemical reactions, the equilibrium expression is a mathematical representation that relates the concentrations of products and reactants at equilibrium. It is specific to each reaction and can be derived from the reaction equation.

For a general reaction:\[ aA + bB \rightleftarrows cC + dD \]the equilibrium constant expression is:\[ K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]

The equilibrium constant expression highlights the ratio of product concentrations to reactant concentrations, each raised to the power of their coefficients from the balanced equation. This quantifies the position of equilibrium; a larger \( K_c \) means more products at equilibrium, and a smaller \( K_c \) means more reactants.

The expression is dimensionless, and factors such as concentration changes or changes in reaction conditions, like pressure or temperature, can affect the value of \( K_c \).
Reaction Equation
A chemical reaction equation is a symbolic representation of a chemical reaction. It shows the reactants transforming into products and is balanced to obey the law of conservation of mass.

For ammonia synthesis, consider two possible reaction equations:
  • \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftarrows 2\text{NH}_3(g)\) and
  • \(\frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g) \rightleftarrows \text{NH}_3(g)\).
The coefficients in these equations are vital. They determine the form of the equilibrium expression and the calculations needed to find the equilibrium constant, \( K_c \).

Balancing reaction equations is crucial for accuracy in any stoichiometric calculations. It ensures that the reaction follows the conservation of mass, where the number of atoms of each element is the same on both sides of the equation. This balance allows for the correct determination of relationships between reactant and product amounts.

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Most popular questions from this chapter

The "boat" form and the "chair" form of cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) interconvert as shown here: $$\underset{k_{-1}}{\stackrel{k_{1}}{\rightleftarrows}}$$ In this representation, the \(\mathrm{H}\) atoms are omitted and a \(\mathrm{C}\) atom is assumed to be at each intersection of two lines (bonds). The conversion is first order in each direction. The activation energy for the chair boat conversion is \(41 \mathrm{~kJ} / \mathrm{mol}\). If the frequency factor is \(1.0 \times 10^{12} \mathrm{~s}^{-1}\), what is \(k_{1}\) at \(298 \mathrm{~K} ?\) The equilibrium constant \(K_{c}\) for the reaction is \(9.83 \times 10^{3}\) at \(298 \mathrm{~K}\).

Consider the following equilibrium process at \(700^{\circ} \mathrm{C}\) : $$ 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{~S}(g) $$ Analysis shows that there are 2.50 moles of \(\mathrm{H}_{2}, 1.35 \times 10^{-5}\) mole of \(\mathrm{S}_{2}\), and 8.70 moles of \(\mathrm{H}_{2} \mathrm{~S}\) present in a 12.0 -L flask. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

At \(1000 \mathrm{~K},\) a sample of pure \(\mathrm{NO}_{2}\) gas decomposes: $$ 2 \mathrm{NO}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The equilibrium constant \(K_{P}\) is 158 . Analysis shows that the partial pressure of \(\mathrm{O}_{2}\) is 0.25 atm at equilibrium. Calculate the pressure of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) in the mixture.

Consider the equilibrium: $$ 2 \mathrm{NOBr}(g) \rightleftarrows 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ If nitrosyl bromide (NOBr) is 34 percent dissociated at \(25^{\circ} \mathrm{C}\) and the total pressure is 0.25 atm, calculate \(K_{P}\) and \(K_{\mathrm{c}}\) for the dissociation at this temperature.

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

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