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Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring formation of reactants? $$ \text { (a) } \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) $$ at \(1300 \mathrm{~K} \quad K_{c}=0.57\) $$ \text { (b) } 2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{C}(s) \text { at } 900 \mathrm{~K} \quad K_{p}=0.0572 $$

Short Answer

Expert verified
Both reactions lie to the left, favoring the formation of reactants.

Step by step solution

01

Understand the Equilibrium Constant

The equilibrium constant, denoted by either \(K_c\) or \(K_p\), indicates the ratio of concentrations of products to reactants at equilibrium. A large \(K\) value (much greater than 1) implies that the equilibrium lies to the right, favoring the formation of products. Conversely, a small \(K\) value (much less than 1) indicates the equilibrium lies to the left, favoring the formation of reactants.
02

Analyze Reaction (a)

For reaction (a), \(K_c = 0.57\). This value is less than 1, suggesting that at equilibrium, there are more reactants than products. Therefore, the equilibrium position favors the formation of reactants, and lies to the left.
03

Analyze Reaction (b)

For reaction (b), \(K_p = 0.0572\). Similar to reaction (a), this value is significantly less than 1. This indicates that the equilibrium position also favors reactants over products, placing the equilibrium to the left.
04

Conclusion

Both reactions have equilibrium constants less than 1. Thus, both reactions favor the formation of reactants, with the equilibrium lying to the left for each reaction.

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, either represented as \( K_c \) for concentrations or \( K_p \) for partial pressures, provides insight into the extent a reaction will proceed to form products under equilibrium conditions.
  • If the equilibrium constant is greater than 1, it signals that the formation of products is favored and the reaction will predominantly lie to the right.
  • Conversely, if the equilibrium constant is less than 1, the formation of reactants is favored, causing the reaction to lie to the left.
To calculate the equilibrium constant, you use the formula that varies slightly depending on whether \( K_c \) or \( K_p \) is being used. For \( K_c \), it is the ratio of the product concentrations to the reactant concentrations, each raised to the power of their respective coefficients in the balanced equation:\[K_c = \frac{[\text{Products}]}{[\text{Reactants}]}\]Similarly, for \( K_p \) which uses pressures, the equation is:\[K_p = \frac{P_{\text{Products}}}{P_{\text{Reactants}}}\]Understanding the magnitude of these constants allows us to predict the position of the equilibrium and hence, ascertain the favored formation of either reactants or products.
Reaction Quotient
The reaction quotient, denoted as \( Q \), offers a snapshot of the reaction's progress at any given moment outside of equilibrium.
  • It is calculated using the same formula as the equilibrium constant, reflecting current concentrations or pressures of reactants and products.
  • If \( Q < K \), the reaction will proceed forward, forming more products as it shifts to the right.
  • If \( Q = K \), the system is already at equilibrium.
  • If \( Q > K \), the reaction will proceed in reverse, forming more reactants as it shifts to the left.
By comparing \( Q \) with \( K \), we can determine which direction the reaction needs to shift to reach equilibrium. This provides a powerful way to predict changes in the concentrations or pressures to reach a state of balance.
Le Chatelier's Principle
Le Chatelier's principle is a guiding concept for understanding how a chemical system at equilibrium responds to external changes. It states that if a system at equilibrium experiences a change in concentration, temperature, or pressure, the system will adjust to counteract the imposed change and re-establish equilibrium.
  • For example, increasing the concentration of a reactant will shift the equilibrium to the right to produce more products.
  • Decreasing the concentration of a reactant will shift equilibrium to the left.
  • An increase in temperature (for endothermic reactions) shifts the equilibrium to the right, favoring products, whereas in exothermic reactions, it shifts to the left.
  • Changes in pressure, affected by a change in volume or the addition of inert gases, can also alter the equilibrium position; the system will favor the side with fewer moles of gas when pressure is increased.
By understanding these principles, we can predict and influence the outcome of reactions in various conditions, making them invaluable in both academic studies and industrial applications.

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Most popular questions from this chapter

A mixture of 0.140 mol of \(\mathrm{NO}, 0.060 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), and 0.260 mol of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a 2.0-L vessel at \(330 \mathrm{~K}\). Assume that the following equilibrium is established: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ At equilibrium \(\left[\mathrm{H}_{2}\right]=0.010 \mathrm{M} .(\mathbf{a})\) Calculate the equilibrium concentrations of \(\mathrm{NO}, \mathrm{N}_{2}\), and \(\mathrm{H}_{2} \mathrm{O} .\) (b) Calculate \(K_{c}\).

Methane, \(\mathrm{CH}_{4}\), reacts with \(\mathrm{I}_{2}\) according to the reaction \(\mathrm{CH}_{4}(g)+\mathrm{I}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{I}(g)+\mathrm{HI}(g) .\) At \(600 \mathrm{~K}, K_{p}\) for this reaction is \(1.95 \times 10^{-4}\). A reaction was set up at 600 \(\mathrm{K}\) with initial partial pressures of methane of \(13.3 \mathrm{kPa}\) and of \(6.67 \mathrm{kPa}\) for \(\mathrm{I}_{2}\). Calculate the pressures, in kPa, of all reactants and products at equilibrium.

At \(900^{\circ} \mathrm{C}, K_{p}=51.2\) for the equilibrium $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ If the pressure of \(\mathrm{NO}(g)\) is half the pressure of \(\mathrm{NOBr}(g)\), what is the equilibrium pressure of \(\mathrm{Br}_{2}(g)\) ?

Write the expressions for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}(g)\) (b) \(\mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SiCl}_{4}(g)\) (c) \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)\) (d) \(\mathrm{O}_{2}(g)+2 \mathrm{CO}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)\) (e) \(\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}^{+}(a q)\) (f) \(\mathrm{Fe}^{2+}(a q)+\mathrm{Ce}^{4+}(a q) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)\) (g) \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)

The following equilibria were measured at \(823 \mathrm{~K}\) : $$ \begin{array}{l} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=67 \\ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=0.14 \end{array} $$ (a) Use these equilibria to calculate the equilibrium constant, \(K_{c}\), for the reaction \(\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)\) \(+\mathrm{CO}_{2}(g)\) at \(823 \mathrm{~K} .(\mathbf{b})\) Based on your answer to part (a), would you say that carbon monoxide is a stronger or weaker reducing agent than \(\mathrm{H}_{2}\) at \(T=823 \mathrm{~K} ?(\mathbf{c})\) If you were to place \(5.00 \mathrm{~g}\) of \(\mathrm{CoO}(s)\) in a sealed tube with a volume of \(250 \mathrm{~mL}\) that contains \(\mathrm{CO}(g)\) at a pressure of \(101.3 \mathrm{kPa}\) and a temperature of \(298 \mathrm{~K},\) what is the concentration of the CO gas? Assume there is no reaction at this temperature and that the CO behaves as an ideal gas (you can neglect the volume of the solid). \((\mathbf{d})\) If the reaction vessel from part (c) is heated to \(823 \mathrm{~K}\) and allowed to come to equilibrium, how much \(\operatorname{CoO}(s)\) remains?

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