Analysis of a reaction mixture showed that it had the composition \(0.417 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}\), \(0.524 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\), and \(0.122 \mathrm{~mol}-\mathrm{L}^{-1} \mathrm{NH}_{3}\) at \(800 \mathrm{~K}\), at which temperature \(K_{c}=0.278\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})=2 \mathrm{NH}_{3}(\mathrm{~g}) \cdot(\mathrm{a}) \mathrm{Calculare}\) the reaction quotient. (b) Is the reaction mixture at equilibrium? (c) If not, is there a tendency to form more reactants or more products?

Short Answer

Expert verified
Q is approximately 0.248436. Since Q < Kc, the reaction mixture is not at equilibrium and there is a tendency to form more products to reach equilibrium.

Step by step solution

01

Write down the balanced equation

The balanced equation for the reaction is given as: \[N_{2}(g) + 3H_{2}(g) = 2NH_{3}(g)\]. It is important to note that the coefficients in the balanced equation will be used as exponents when calculating the reaction quotient.
02

Write down the expression for the reaction quotient (Q)

The reaction quotient (Q) for a reaction is given by the formula: \[Q = \frac{([\text{Products}])^\text{coefficients}}{([\text{Reactants}])^\text{coefficients}}\]. In this case, \[Q = \frac{([NH_{3}])^2}{([N_{2}])[H_{2}]^3}\].
03

Plug in the concentrations into the Q expression

Using the given concentrations, we plug them into the Q expression: \[Q = \frac{(0.122)^2}{(0.417)(0.524)^3}\].
04

Calculate the reaction quotient (Q)

Perform the calculations: \[Q = \frac{(0.122)^2}{(0.417)(0.524)^3} = \frac{0.014884}{(0.417)(0.143551424)} = \frac{0.014884}{0.059901842688}\] which yields \[Q \approx 0.248436\].
05

Compare Q with the equilibrium constant (Kc)

Compare the calculated Q value with the given equilibrium constant (Kc) to determine if the reaction is at equilibrium. The given Kc value is 0.278.
06

Determine the direction of the reaction

If Q < Kc, the reaction will proceed in the forward direction to reach equilibrium, forming more products. If Q > Kc, the reaction will proceed in the reverse direction to reach equilibrium, forming more reactants.
07

Conclusion

Since the calculated Q value of 0.248436 is less than the given Kc value of 0.278, the reaction will proceed in the forward direction to reach equilibrium, forming more products.

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Key Concepts

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

Chemical Equilibrium
When a chemical reaction occurs, the reactants transform into products, but under certain conditions, the products can also revert back to reactants. Chemical equilibrium is reached when the rates of the forward and reverse reactions are equal, leading to constant concentrations of both reactants and products within the system over time.

This state doesn't imply that the reactants and products are in equal concentrations, but rather that their concentrations have stabilized in a fixed proportion. Even though individual molecules are continuously reacting, there is no net change in the concentration of the species involved.

In our exercise, to determine if the reaction mixture is at equilibrium, we need to calculate the reaction quotient (Q) and compare it with the equilibrium constant (Kc). If Q equals Kc, the system is at equilibrium. If not, it indicates which direction the reaction must proceed to reach equilibrium.
Equilibrium Constant (Kc)
The equilibrium constant (Kc) is a numerical value that relates the concentrations of reactants and products of a reaction at equilibrium at a given temperature. The formula for Kc is derived based on the balanced chemical equation. For a generic reaction: \(aA + bB \leftrightarrow cC + dD\), the equilibrium constant expression would be: \[Kc = \frac{[C]^c[D]^d}{[A]^a[B]^b}\].

Several important points should be noted about Kc: it is dimensionless, it applies only to reactions at equilibrium, and it is dependent on temperature. A Kc value greater than one suggests products are favored at equilibrium, whereas a Kc less than one suggests that reactants are favored. In the provided exercise, the Kc is given as 0.278, meaning at 800 K, the reaction moderately favors the formation of reactants.
Balanced Chemical Equation
A balanced chemical equation faithfully represents the law of conservation of mass, indicating that the number of atoms of each element is conserved in the course of a chemical reaction. It provides the stoichiometry of the reaction, which is essential for calculating the reaction quotient (Q) and the equilibrium constant (Kc).

In each balanced chemical equation, coefficient ratios express the smallest whole-number ratios of moles of reactants and products involved in the reaction. Just as the balanced equation \(N_{2}(g) + 3H_{2}(g) = 2NH_{3}(g)\) indicates that one mole of nitrogen gas reacts with three moles of hydrogen gas to produce two moles of ammonia gas, the coefficients (1, 3, and 2) are used as exponents for their respective species in the equilibrium expression.

A correctly balanced chemical equation is critical for predicting the outcome of a reaction, designing chemical processes, and calculating yields.

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Most popular questions from this chapter

Calculate the standard reaction free energy of each of the following reactions: (a) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}), K=41\) ar \(400 \mathrm{~K}\) (b) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}), K=3.0 \times 10^{4}\) at \(700 \mathrm{~K}\)

A mixture consisting of \(1.1 \mathrm{mmol} \mathrm{SO}_{2}\) and \(2.2 \mathrm{mmol} \mathrm{O}_{2}\) in a \(250-\mathrm{mL}\) container was heated to \(500 \mathrm{~K}\) and allowed to reach equilibrium. Will more sulfur trioxide be formed if that equilibrium mixture is cooled to \(25^{\circ} \mathrm{C}\) ? For the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\) \(\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g}), \mathrm{K}=2.5 \times 10^{10}\) at \(500 \mathrm{~K}\) and \(4.0 \times 10^{24}\) at \(25^{\circ} \mathrm{C}\).

A mixture consisting of \(1.000 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) and \(1.000 \mathrm{~mol} \mathrm{CO}(\mathrm{g})\) is placed in a \(10.00-\mathrm{L}\) reaction vessel at \(800 \mathrm{~K}\). At equilibrium, \(0.665 \mathrm{~mol} \mathrm{CO}_{2}(\mathrm{~g})\) is present as a result of the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})=\) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\). What are (a) the equilibrium concentrations for all substances and (b) the value of \(K_{e}\) ?

The equilibrium constant \(K=3.5 \times 10^{4}\) for the reaction \(\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})=\mathrm{PCl}_{5}(\mathrm{~g})\) at \(760^{\circ} \mathrm{C}\). At equilibrium, the partial pressure of \(\mathrm{PCl}_{5}\) was \(2.2 \times\) \(10^{4}\) bar and that of \(\mathrm{PCl}_{3}\) was \(1.33\) bar. What was the equilibrium partial pressure of \(\mathrm{Cl}_{2}\) ?

Predict whether each of the following equilibria will shift toward products or reactants with a temperature increase. (a) \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g})\), \(\Delta H^{\circ}=+206 \mathrm{~kJ}\) (b) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})=\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\), \(\Delta H^{2}=-41 \mathrm{~kJ}\) (c) \(2 \mathrm{SO}_{2}\) (g) \(+\mathrm{O}_{2}\) (g) \(=2 \mathrm{SO}_{3}\) (g), \(\Delta H^{\circ}=-198 \mathrm{~kJ}\)

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