Given the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)\), consider the following situations: i. You have \(1.3 M \mathrm{~A}\) and \(0.8 \mathrm{M}\) B initially. ii. You have \(1.3 M \mathrm{~A}, 0.8 \mathrm{M} \mathrm{B}\), and \(0.2 \mathrm{M} \mathrm{C}\) initially. iii. You have \(2.0 M \mathrm{~A}\) and \(0.8 M \mathrm{~B}\) initially. Order the preceding situations in terms of increasing equilibrium concentration of D. Explain your order. Then give the order in terms of increasing equilibrium concentration of \(\mathrm{B}\) and explain.

Short Answer

Expert verified
The order for increasing equilibrium concentration of D is Situation i < Situation iii < Situation ii, and the order for increasing equilibrium concentration of B is Situation ii < Situation i < Situation iii. This is because, in all cases, the reaction proceeds in the forward direction and more D will be formed. Situation ii has more moles of B available per mole of A, resulting in more D being formed at equilibrium. As for B, it is consumed more in Situation ii, resulting in a lower equilibrium concentration.

Step by step solution

01

Determine the initial reaction quotient, Q, for each situation

To find out the behavior of the system in each situation, we need to determine the initial reaction quotient, Q. Recall that: \[Q = \frac{[\mathrm{C}] [\mathrm{D}]}{[\mathrm{A}] [\mathrm{B}]}\] For Situation i: Q₁ = \(\frac{0 \cdot 0}{1.3 \cdot 0.8} = 0\) For Situation ii: Q₂ = \(\frac{0.2 \cdot 0}{1.3 \cdot 0.8} = 0\) For Situation iii: Q₃ = \(\frac{0 \cdot 0}{2.0 \cdot 0.8} = 0\)
02

Analyze the initial reaction quotient

Since this is a generic reaction, we don't know the exact value of the equilibrium constant K. However, we can still analyze the system based on the initial reaction quotient, Q. As Q < K for all the situations (since all Q’s are zero), the reaction will proceed in the forward direction until equilibrium is reached.
03

Rank situations in terms of increasing equilibrium concentration of D

To rank the situations in terms of increasing equilibrium concentration of D, we need to analyze how much D will be generated from the initial concentrations of A and B in each case. In all cases, since the reaction will proceed forward, more D will be formed. Also, in all situations the limiting reactant is B. Situation i: [(0.8 mol/L B) / (2 mol/L A)] = 0.4 Situation ii: [(0.8 mol/L B) / (1.3 mol/L A)] = 0.615 Situation iii: [(0.8 mol/L B) / (2 mol/L A)] = 0.4 We can see that Situation ii will result in more D being formed at equilibrium, as there's more B available per mole of A. Thus, the order of increasing equilibrium concentration of D is: Situation i < Situation iii < Situation ii
04

Rank situations in terms of increasing equilibrium concentration of B

To rank the situations in terms of increasing equilibrium concentration of B, we need to analyze how much B will be consumed in each case. Since limiting reactant is B and it will be used up more in Situation ii, it will have the least concentration at equilibrium. Therefore, the order of increasing equilibrium concentration of B is: Situation ii < Situation i < Situation iii In summary, the order for increasing equilibrium concentration of D is Situation i < Situation iii < Situation ii, and the order for increasing equilibrium concentration of B is Situation ii < Situation i < Situation iii.

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Key Concepts

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

Reaction Quotient (Q)
When studying chemical equilibrium, the reaction quotient (Q) is a vital concept that tells us the direction in which a reaction will proceed before it reaches equilibrium. Specifically, it is a ratio of the concentrations of products to reactants at any point in time and is given by the equation similar to the equilibrium constant (K), but Q can be calculated at any stage of the reaction, not just at equilibrium.

For the given reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)\), the reaction quotient is calculated as:\[Q = \frac{[\mathrm{C}] [\mathrm{D}]}{[\mathrm{A}] [\mathrm{B}]}\].

In the exercise provided, we determined the initial Q for various initial concentrations of reactants to predict the reactions' behaviors. If Q is less than the equilibrium constant K, the forward reaction is favored, and reactants will be converted into products until equilibrium is achieved. Contrarily, if Q is greater than K, the reverse reaction is favored, and products will be converted back to reactants. With Q equal to zero in all scenarios, the reactions proceed towards forming more products.
Equilibrium Concentration
The equilibrium concentration of a substance is the concentration of that substance when the reaction has reached a state of balance and no further changes occur in the concentrations of reactants and products. It's a key aspect to understand since all calculations for equilibrium involve concentrations of reactants and products at this static point.

For the exercise, after determining that the reactions would proceed forward based on the initial Q values, we analyzed the equilibrium concentration of D by considering the amounts of A and B present initially. Knowing that the reaction moves forward until equilibrium is reached, we can infer that the more of the limiting reactant available, the more product can be formed. Hence, in situations with different initial amounts of reactants, the equilibrium concentrations will vary and must be arranged according to the availability of the limiting reactant, which determines the potential product formation.
Limiting Reactant
The concept of a limiting reactant is essential in stoichiometry and chemical equilibrium. It is the reactant that is entirely consumed first during a chemical reaction, which limits the amount of product that can be formed and determines when the reaction will stop.

In our exercise, by comparing the initial molar ratios and knowing that B is the limiting reactant, we can predict the extent of the reaction. For instance, although A is present in excess in all situations, only a specific amount of D can be produced based on the initial quantity of B. Consequently, we see that when B is present in a greater amount relative to A (as in Situation ii), more of D can be formed. This demonstrates the limiting nature of reactant B, as it dictates the maximum yield of product in each scenario.

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Most popular questions from this chapter

At high temperatures, elemental nitrogen and oxygen react with each other to form nitrogen monoxide: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be \(\left[\mathrm{N}_{2}\right]=\) \(0.041 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.0078 M\), and \([\mathrm{NO}]=4.7 \times 10^{-4} M .\) Calcu- late the value of \(K\) for the reaction.

Consider an equilibrium mixture of four chemicals \((\mathrm{A}, \mathrm{B}, \mathrm{C},\), and \(\mathrm{D}\), all gases) reacting in a closed flask according to the equation: $$ \mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) $$ a. You add more \(\mathrm{A}\) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. b. You have the original setup at equilibrium, and you add more \(\mathrm{D}\) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ Calculate the concentrations of all species at equilibrium for each of the following original mixtures. a. \(2.0\) moles of pure \(\mathrm{NOCl}\) in a \(2.0-\mathrm{L}\) flask b. \(1.0\) mole of \(\mathrm{NOCl}\) and \(1.0 \mathrm{~mole}\) of \(\mathrm{NO}\) in a \(1.0-\mathrm{L}\) flask c. \(2.0\) moles of \(\mathrm{NOCl}\) and \(1.0 \mathrm{~mole}\) of \(\mathrm{Cl}_{2}\) in a \(1.0-\mathrm{L}\) flask

Hydrogen for use in ammonia production is produced by the reaction $$ \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \stackrel{\mathrm{Nicatalyst}}{750^{\circ} \mathrm{C}} \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) $$ What will happen to a reaction mixture at equilibrium if a. \(\mathrm{H}_{2} \mathrm{O}(g)\) is removed? b. the temperature is increased (the reaction is endothermic)? c. an inert gas is added to a rigid reaction container? d. \(\mathrm{CO}(g)\) is removed? e. the volume of the container is tripled?

The value of the equilibrium constant \(K\) depends on which of the following (more than one answer may be correct)? a. the initial concentrations of the reactants b. the initial concentrations of the products c. the temperature of the system d. the nature of the reactants and products Explain.

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