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Analysis of a reaction mixture showed that it had the composition \(0.624 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}, 0.315 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\), and \(0.222 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{NH}_{3}\) at \(800 . \mathrm{K}\), at which temperature \(K_{c}=0.278\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}\) (g). (a) Calculate the reaction quotient \(Q_{c}\). (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
\(Q_c \approx 0.153\); since \(Q_c < K_c\), the reaction will proceed in the forward direction, forming more products.

Step by step solution

01

Write the Balanced Chemical Equation

Write down the balanced chemical equation for the reaction and identify the reactants and products: \[ \mathrm{N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)} \]
02

Write the Expression for the Reaction Quotient, \(Q_c\)

Write the expression for the reaction quotient \(Q_c\) based on the chemical equation: \[ Q_c = \frac{[\mathrm{NH_{3}}]^2}{[\mathrm{N_{2}}][\mathrm{H_{2}}]^3} \]
03

Calculate the Reaction Quotient, \(Q_c\)

Substitute the given concentrations into the \(Q_c\) expression and calculate its value: \[ Q_c = \frac{(0.222)^2}{(0.624)(0.315)^3} \]
04

Compare \(Q_c\) to the Equilibrium Constant, \(K_c\)

Compare the calculated \(Q_c\) value to the given equilibrium constant \(K_c\) to determine if the reaction is at equilibrium: If \(Q_c = K_c\), the system is at equilibrium; if \(Q_c < K_c\), the reaction moves forward to form more products; if \(Q_c > K_c\), the reaction moves backward to form more reactants.
05

Analyze the Reaction Direction

Decide on the direction the reaction will proceed based on the comparison of \(Q_c\) and \(K_c\) from Step 4 to answer whether there is a tendency to form more reactants or more products.

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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 \textbf{Qc}, is critical when analyzing chemical reactions in a non-equilibrium state. It predicts the direction in which a chemical reaction is likely to proceed to reach equilibrium. The reaction quotient is calculated in a similar manner to the equilibrium constant, but with the concentrations of reactants and products at any point in time, not just at equilibrium.

\textbf{How to calculate Qc:} For the generic reaction \(aA + bB \rightleftharpoons cC + dD\), where A and B are reactants and C and D are products, the reaction quotient is given by:\[Q_c = \frac{\text{[C]}^c \times \text{[D]}^d}{\text{[A]}^a \times \text{[B]}^b}\]Here, \textbf{[A]}, \textbf{[B]}, \textbf{[C]}, and \textbf{[D]} are the molar concentrations of the respective chemicals at a given moment, and \(a, b, c,\) and \(d\) are the stoichiometric coefficients from the balanced chemical equation.

\textbf{Significance of Qc:} By comparing \(Q_c\) to the equilibrium constant \(K_c\), one can predict whether a reaction will proceed to form more reactants or products. This prediction is vital for chemists to understand how conditions such as concentration and temperature affect the reaction progress.
Equilibrium Constant (Kc)
The equilibrium constant, represented as \textbf{Kc}, embodies the ratio of product concentrations to reactant concentrations for a chemical reaction at equilibrium. It is an essential figure that indicates how far a reaction goes to reach a state of balance and is only affected by temperature.\textbf{Expression for Kc:} Based on a balanced chemical equation, say \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant is expressed as:\[K_c = \frac{\text{[C]}^c \times \text{[D]}^d}{\text{[A]}^a \times \text{[B]}^b}\]In this expression,\[\] refers to the concentration of the respective components at equilibrium, and the exponents are the stoichiometric coefficients.

\textbf{Interpreting Kc:} A large \(K_c\) value implies a product-favored reaction, where the reactants are mostly converted into products when equilibrium is reached. Conversely, a small \(K_c\) indicates a reactant-favored reaction where only a small fraction of reactants transform into products.
Balanced Chemical Equation
A \textbf{balanced chemical equation} is fundamental in chemistry because it ensures that the Law of Conservation of Mass is obeyed during a chemical reaction. It indicates that the number of atoms for each element is the same on both sides of the equation.\textbf{Creating a balanced equation involves:}
  • Writing down the correct formulas for all reactants and products.
  • Adjusting coefficients, the numbers in front of formulas, to ensure that the same number of each type of atom appears on both sides of the equation.
For instance, in the reaction example \(\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}\), '1' is the coefficient for nitrogen (\textbf{\(\mathrm{N_2}\)}), '3' for hydrogen (\textbf{\(\mathrm{H_2}\)}), and '2' for ammonia (\textbf{\(\mathrm{NH_3}\)}), illustrating that for every molecule of nitrogen and three molecules of hydrogen that react, two molecules of ammonia are produced. This balance is crucial in correctly calculating the reaction quotient and the equilibrium constant.
Chemical Reaction Direction
The \textbf{chemical reaction direction} is indicative of whether a reaction will proceed to form more reactants or products at any given point relative to the state of equilibrium. This directionality is predicted by comparing the reaction quotient \(Q_c\) to the equilibrium constant \(K_c\).\textbf{Determining the direction:} If \(Q_c < K_c\), the system is not yet at equilibrium, and the concentration of products is lower than it should be at equilibrium. As a result, the reaction will ‘move forward’ or proceed in the forward direction to generate more products and fewer reactants. On the other hand, if \(Q_c > K_c\), there are more products present than at equilibrium, so the reaction will ‘move backward,’ converting products back into reactants to achieve equilibrium. When \(Q_c = K_c\), the reaction is at equilibrium, and no net change in the concentrations of reactants or products occurs.It is this understanding of reaction direction that allows chemists to manipulate conditions to shift the equilibrium in favor of desired products or reactants, a principle used in many industrial and laboratory chemical processes.

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Most popular questions from this chapter

\mathrm{~A} 0.10-\mathrm{mol}\( sample of \)\mathrm{H}_{2}(\mathrm{~g})\( and a \)0.10-\mathrm{mol}\( sample of \)\mathrm{Br}_{2}(\mathrm{~g})\( are placed into a \)2.0\( - L container. The reaction \)\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons\( \)2 \mathrm{HBr}(\mathrm{g})\( is then allowed to come to equilibrium. A \)0.20\(-mol sample of HBr is placed into a second 2.0-L sealed container at the same temperature and allowed to reach equilibrium with \)\mathrm{H}_{2}\( and \)\mathrm{Br}_{2}\(. Which of the following will be different in the two containers at equilibrium? Which will be the same? (a) Amount of \)\mathrm{Br}_{2}\(; (b) concentration of \)\mathrm{H}_{2}\(; (c) the ratio \)[\mathrm{HBr}] /\left[\mathrm{H}_{2}\right]\left[\mathrm{Br}_{2}\right]\(; (d) the ratio \)[\mathrm{HBr}] /\left[\mathrm{Br}_{2}\right]\(; (e) the ratio \)\left[\mathrm{HBr}^{2} /\left[\mathrm{H}_{2}\right]\left[\mathrm{Br}_{2}\right]\right.$; (f) the total pressure in the container. Explain each of your answers.

Write the reaction quotient \(Q\) for (a) \(2 \mathrm{BCl}_{3}\) (g) \(+2 \mathrm{Hg}(\mathrm{l}) \rightarrow \mathrm{B}_{2} \mathrm{Cl}_{4}\) (s) \(+\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) (s) (b) \(\mathrm{P}_{4} \mathrm{~S}_{10}\) (s) \(+16 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})+10 \mathrm{H}_{2} \mathrm{~S}(\mathrm{aq})\) (c) \(\mathrm{Br}_{2}(\mathrm{~g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{BrF}_{3}(\mathrm{~g})\)

Consider the equilibrium \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\) \(\mathrm{H}_{2}(\mathrm{~g})\). (a) If the partial pressure of \(\mathrm{CO}_{2}\) is increased, what happens to the partial pressure of \(\mathrm{H}_{2}\) ? (b) If the partial pressure of \(\mathrm{CO}\) is decreased, what happens to the partial pressure of \(\mathrm{CO}_{2}\) ? (c) If the concentration of \(\mathrm{CO}\) is increased, what happens to the concentration of \(\mathrm{H}_{2}\) ? (d) If the concentration of \(\mathrm{H}_{2} \mathrm{O}\) is decreased, what happens to the equilibrium constant for the reaction?

Determine whether the following statements are true or false. If false, explain why. (a) In an equilibrium reaction, the reverse reaction begins only when all reactants have been converted to products. (b) The equilibrium concentrations will be the same whether one starts with pure reactants or pure products. (c) The rates of the forward and reverse reactions are the same at equilibrium. (d) If the Gibbs free energy is greater than the standard Gibbs free energy of reaction, the reaction proceeds forward to equilibrium.

Cyclohexane \((C)\) and methylcyclopentane \((M)\) are isomers with the chemical formula \(\mathrm{C}_{6} \mathrm{H}_{12}\). The equilibrium constant for the rearrangement \(\mathrm{C} \rightleftharpoons \mathrm{M}\) in solution is \(0.140\) at \(25^{\circ} \mathrm{C}\). (a) \(\mathrm{A}\) solution of \(0.0200 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) cyclohexane and \(0.100 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) methylcyclopentane is prepared. Is the system at equilibrium? If not, will it will form more reactants or more products? (b) What are the concentrations of cyclohexane and methylcyclohexane at equilibrium? (c) If the temperature is raised to \(50 .{ }^{\circ} \mathrm{C}\), the concentration of cyclohexane becomes \(0.100 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) when equilibrium is re-established. Calculate the new equilibrium constant. (d) Is the reaction exothermic or endothermic at \(25^{\circ} \mathrm{C}\) ? Explain your conclusion.

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